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<h2 class="hd hd-2 unit-title">The Theorem</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> With three warm-up cases under our belts, we’re ready to tackle the Banach-Tarski Theorem. (I’ll skip some of the technical details. If you’d like a more rigorous version of the proof, have a look at some of the resources listed in Lecture 8.4.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><strong><span style="font-size: 12pt;">Cayley on a Ball</span></strong></span></p>
<p><span style="font-family: 'book antiqua', palatino;">The first step of the proof is to construct a <em>modified</em> version of the Cayley Graph—a version of the graph that is wrapped around the surface of a ball, instead of living on a flat plane.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The “center” of our graph is an arbitrary point [mathjaxinline]c [/mathjaxinline] on the surface of our ball. As in the case of the original graph, our modified graph consists of an infinite number of “Cayley Paths”. As before, each Cayley Path starts from [mathjaxinline] c[/mathjaxinline], and is the result of taking a finite sequence of steps. As before, there are four different kinds of steps: “up”, “down”, “right” and “left”. And, as before, one is not allowed to follow a step with its inverse. But there is an important difference. In the original Cayley graph, a “step” was the result of traversing a certain distance on the plane. In our modified Cayley graph, in contrast, a “step” is result of performing a <em>rotation</em> on the sphere.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">To see what the relevant rotations consist in, imagine that you’re holding the ball in your hand, and consider the following axes:</span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">The \(x\)-axis runs from your right to your left through the center of the ball. </span></li>
<li><span style="font-family: book antiqua, palatino;">The \(y\)-axis runs is orthogonal to the \(x\)-axis. It runs from the wall in front of you to the wall in behind of you, through the center of the ball.</span></li>
<li><span style="font-family: book antiqua, palatino;">The the \(z\)-axis is orthogonal to the other two. It runs from the ground to the sky, through the center of the ball. </span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">For a given angle \(\theta\) (which will be chosen carefully---more about this later), we will be concerned with the following rotations:</span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">An "up'' rotation is a counterclockwise rotation of \(\theta\) degrees about the \(x\) axis. (When you're holding the ball in front of you, you perform this rotation by rotating the ball from bottom to top.)</span></li>
<li><span style="font-family: book antiqua, palatino;">A "down'' rotation is a clockwise rotation of \(\theta\) degrees about the \(x\) axis. (When you're holding the ball in front of you, you perform this rotation by rotating the ball from top to bottom.)</span></li>
<li><span style="font-family: book antiqua, palatino;">A "right'' rotation is a counterclockwise rotation of \(\theta\) degrees about the \(z\) axis. (When you're holding the ball in front of you, you perform this rotation by rotating the ball from left to right.)</span></li>
<li><span style="font-family: book antiqua, palatino;">A "left'' rotation is a clockwise rotation of \(\theta\) degrees about the \(z\) axis. (When you're holding the ball in front of you, you perform this rotation by rotating the ball from right to left.)</span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">To each such rotation \(\rho\) corresponds a function \(f_\rho\), which takes each point \(p\) on the surface of the ball to the point on the surface of the ball whose current location (relative to an external reference frame) would come to be occupied by \(p\) were rotation \(\rho\) to be performed. For instance, \(f_\text{right}\) is the function that takes each point \(p\) to the point that is currently \(\theta\) degrees East of \(p\).</span><br /><br /><span style="font-family: book antiqua, palatino;">Now return to our modified version of the Cayley graph. I said earlier that each path of the graph is the result of taking a sequence of "up'', "down'', "right'' and "left'' steps (without ever following a step with its inverse). I am now in a position to explain what it means to "follow'' one of these paths on the surface of the ball. Consider the following example of a Cayley Path:</span><br /><span style="font-family: book antiqua, palatino;">\[\langle\text{up}, \text{right}, \text{up}, \text{up}, \text{left}, \text{down}\rangle\]</span><br /><span style="font-family: book antiqua, palatino;">In the original version of the Cayley Graph, one follows this path by starting at the graph's center and moving through the plane: first upwards, then rightwards, then upwards again, and so forth. In our modified version of the Cayley Graph, one "follows'' the path by starting at the graph's "central'' point \(c\) and applying each of the rotations in the path, in reverse order. Accordingly, the "endpoint'' of the path above is:</span><br /><span style="font-family: book antiqua, palatino;">\[f_{\text{up}}(f_{\text{right}}(f_{\text{up}}(f_{\text{up}}(f_{\text{left}}(f_{\text{down}}(c))))))\]</span><br /><span style="font-family: book antiqua, palatino;">In general, the "endpoint" of Cayley Path \(\langle\rho_1,\rho_2,\dots,\rho_n\rangle\) is the point \(f_{\rho_1}(f_{\rho_2}(\dots f_{\rho_n}(c)\dots))\).</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">Sydney Gibson, who was a star student in the residential version of <em>Paradox and Infinity</em> I taught last year at MIT, has been working on a nice tool for visualizing our rotations, which you can find <a href="http://web.mit.edu/arayo/www/Cayley/viz.html" target="[object Object]">here</a>. (Use your keyboard arrows to take Cayley Steps and your mouse to change your orientation with respect to the ball.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Note: this is a \(\beta\) version of the visualization. Sydney is still working on the code, but she graciously agreed to make it available in its current state to make sure you all had a chance to use it while you're still taking the class.</span></p>
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<span style="font-family: 'book antiqua', palatino;">By representing rotations as matrices, give an analytic characterization of each of our four rotations (up, down, right, left) and the corresponding functions (\(f_{\text{up}}, f_{\text{down}}, f_{\text{right}}, f_{\text{left}}\)).</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Note:</i> This is a difficult exercise, unless you're comfortable with trigonometry and matrix multiplication. If you're not, you might like to use the answer to learn more about how to work out the details of our rotations.</span>
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<h2 class="hd hd-2 unit-title">The key property of Rotations</h2>
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<p><span style="font-family: book antiqua, palatino;">When the rotation \(\theta\) is chosen properly we get an important result. (Let us assume, for concreteness, that \(\theta = \text{arccos}(1/3) \approx 70.53^\circ\), since that turns out to be one of the choices of rotation angle that works.) The result is that <em>different Cayley Paths always have different endpoints</em>. In other words:</span></p>
<p style="padding-left: 30px;"><span style="font-family: book antiqua, palatino;"><strong>Key Property</strong><br />If \(\langle\rho_1,\rho_2,\dots,\rho_n\rangle\) and \(\langle\sigma_1,\sigma_2,\dots,\sigma_m\rangle\) are distinct Cayley Paths, then \[f_{\rho_1}(f_{\rho_2}(\dots f_{\rho_n}(c)\dots)) \neq f_{\sigma_1}(f_{\sigma_2}(\dots f_{\sigma_m}(c)\dots))\] </span></p>
<p><span style="font-family: book antiqua, palatino;">(Annoyingly, the Key Property holds for <em>almost</em> every choice of \(c\) but not quite every choice. I'll come back to this <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/74332593b7d7439ba2e645ba39eacc0c" target="[object Object]">later</a>. Until then, I'll simply assume that \(c\) has been chosen property.)<br /></span><br /><span style="font-family: book antiqua, palatino;">The Key Property is, in effect, a sophisticated version of a point we came across in </span><span style="font-family: book antiqua, palatino;"><span style="font-family: book antiqua, palatino;"><span style="font-family: book antiqua, palatino;"> <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/96bae303e758441cb2f3a0a7bbd15423" target="[object Object]">Warm-Up Case 2</a>, </span></span> when we saw that one can never return to one's starting point by traveling around the unit circle in steps of one unit. The proof is long and tedious, so I won't go through it here, but it is spelled out in the second of the two readings I list at the end of this lecture. <br /><br />I will, however, illustrate the Key Property with a couple of examples. To setup our examples, we'll need to start by setting up a system of geographical coordinates: a system of lines of latitude and longitude. We will let the "North Pole'' and the "South Pole'' of our ball be the top-most and bottom-most points of the ball, respectively. A line of latitude (or a "parallel'') is a circle on the ball's surface whose points are at a uniform distance from the ball's North Pole. The ball's "equator'' is its largest parallel. A line of longitude (or a "meridian'') is a great circle that crosses the poles. Finally, the ball's "Prime Meridian'' is an arbitrarily selected meridian; for the sake of concreteness, I'll count the meridian that crosses the point on the ball closest to you as the Prime Meridian.<br /><br />We will think of this as an "external'' coordinate system. In other words: our coordinates do not change their locations with respect to the room as you rotate the ball. You can think of them as projected onto the ball by lasers, rather than painted on the ball. So, for instance, the North and South Poles of our ball will always be the top-most and bottom-most points of the ball, regardless of how the ball moves about as we perform particular rotations. <br /><br />Let us think of the point which is currently at the intersection of the ball's equator and its Prime Meridian as our "central'' point \(c\). We will mark its location on the ball with a pencil, and see what happens to our mark as we perform particular rotations. For the sake of concreteness, let us follow the Cayley Path \(\langle\text{right},\text{up}\rangle\). <br /><br /></span></p>
<p><span style="font-family: book antiqua, palatino;"></span><br /><span style="font-family: book antiqua, palatino;"></span><span style="font-family: book antiqua, palatino;"></span></p>
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<p><strong><span style="font-family: book antiqua, palatino;">First Example: \(<\text{right},\text{up}>\)</span></strong></p>
<p><span style="font-family: book antiqua, palatino;"><span style="font-family: book antiqua, palatino;">Let us follow the Cayley Path \(<\text{right},\text{up}>\). </span>One follows a path by applying each of the rotations in the path, in reverse order. So we start by performing an ``up'' rotation. In other words: we rotate the ball \(\theta\) degrees about the \(x\)-axis (which runs from your left to your right, through the center of the ball). </span></p>
<p><span style="font-family: book antiqua, palatino;"> The effect of this rotation is that the mark goes from being located at (\(0^\circ,0^\circ\)) (which is the intersection of the Prime Meridian and the equator), to being located at (\(\theta^\circ \text{ North},0^\circ\)) (which is a latitude of \(\theta^\circ\) North, on the Prime Meridian). On Earth, this corresponds to the location you would get to if you traveled north from London until you reached the latitude of northern Norway. </span></p>
<p><span style="font-family: book antiqua, palatino;">Next, we perform a \(right\) rotation. In other words, we rotate the ball \(\theta\) degrees about the \(z\)-axis (which runs from the ground to the sky, through the center of the ball). This causes the pencil mark to travel East on the \(\theta\)th parallel until it reaches (\(\theta^\circ\) North, \(\theta^\circ\) East). On Earth, this corresponds to the location of the Yamal peninsula, in northern Siberia.</span><br /><br /><span style="font-family: book antiqua, palatino;">Note that even though our "up'' and "right'' rotations were both rotations of \(\theta\) degrees, they did not cause our mark to travel the same distance across the surface of the ball. In general, the distance traveled by our mark during a particular rotation depends on its distance from the <em>poles of rotation</em> (i.e. the points at which the axes of rotation intersect the surface of the ball). </span></p>
<p><span style="font-family: book antiqua, palatino;">If our mark is located close to a pole of rotation, it will travel around a small circle (specifically: the unique circle on the surface of the ball that contains our mark and is centered on the pole of rotation), and it will cover a distance corresponding to \(\frac{\theta}{360}\) of the circumference of that circle. </span></p>
<p><span style="font-family: book antiqua, palatino;">If, on the other hand, our mark is equidistant between the two poles of rotation, it will travel around the great circle of points equidistant from those poles, and it will cover a distance corresponding to \(\frac{\theta}{360}\) of the circumference of that larger circle. </span><br /><br /><span style="font-family: book antiqua, palatino;">Now go back to our example. </span></p>
<p><span style="font-family: book antiqua, palatino;">When you perform the initial "up'' rotation, the mark is exactly equidistant between the poles of rotation: it is at \((0^\circ,0^\circ)\) and the poles of an "up'' rotation (which is a rotation about the \(x\)-axis) are at \((0^\circ,90^\circ \text{ East})\) and \((0^\circ,90^\circ \text{ West})\). So our mark travels a distance of \(\frac{\theta}{360}\) of the circumference of a great circle. </span></p>
<p><span style="font-family: book antiqua, palatino;">In contrast, when you perform the "right'' rotation the mark is close to one of the axes of rotation: it is at \((\theta^\circ \text{ North},0^\circ)\) and the poles of a "right'' rotation (which is a rotation about the \(z\)-axis) are at the ball's North and South poles. (Recall that \(\theta \approx 70.53^\circ\).) So our mark travels a relatively short distance: \(\frac{\theta}{360}\) of the circumference of the ball's \(\theta\)th parallel. Because \(\theta = \text{arccos}(1/3)\), the radius of this parallel is a third of the radius of the ball, so the distance traveled by our mark will be a third of the distance traveled in the first rotation.</span></p>
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<p><strong><span style="font-family: book antiqua, palatino;">Second Example: \(<\text{up},\text{right}>\)</span></strong></p>
<p><span style="font-family: book antiqua, palatino;">We have considered result of following the Cayley Path \(<\text{right},\text{up}>\). Let us compare this with the result of following the Cayley Path \(<\text{up},\text{right}>\). </span></p>
<p><span style="font-family: book antiqua, palatino;">As usual, we proceed by applying each of the rotations in the path, in reverse order. Our pencil mark starts out at \((0^\circ,0^\circ)\), which is equidistant between the balls North and South poles. So a "right'' rotation will cause our mark to travel a distance of \(\frac{\theta}{360}\) of the circumference a great circle, and end up at \((0^\circ,\theta^\circ \text{ East})\), which on Earth corresponds to a location in the middle of the Indian Ocean, west of the Maldives. </span></p>
<p><span style="font-family: book antiqua, palatino;">This point is close to \((0^\circ,90^\circ \text{ East})\), which is one of the poles of an "up'' rotation. So applying such a rotation will cause our mark to travel a relatively short distance and end up close to \((18.32^\circ \text{ North}, 83.28^\circ \text{ East})\), which on Earth corresponds to a location near the village of Garbham on the east coast of India. (As before, the distance traveled in the second rotation will be a third of the distance traveled in the first rotation.)</span><br /><br /><span style="font-family: book antiqua, palatino;">The lesson of our examples is that the Cayley Paths \(<\)right, up\(>\) and \(<\)up, right\(>\) have different endpoints. In the first case, our mark starts out traveling a long distance "up'' and then travels a short distance "right''; in the second case, our mark starts out traveling a long distance "right'' and then travels a short distance "up''. </span></p>
<p><span style="font-family: book antiqua, palatino;">This illustrates the key property of our rotations: as long as \(c\) is chosen properly, different Cayley Paths will always have different endpoints. <br /></span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><strong>Warning:</strong> in the video below I use a coordinate system that rotates as I rotate the ball rather than an external coordinate system, which would have remained fixed with respect to the room as I performed my rotations. As a result, I'm forced to reverse the rotations to get the right results. In other words: I simulate the result of a right rotation by rotating <em>left</em> and the result of an up rotation by rotating <em>down</em>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">I hope this isn't too confusing... My request for a laser-projected coordinated system was turned down without comment. </span></p>
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<h3 class="hd hd-2">Video Review: The Key Property of Rotations</h3>
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<span style="font-family: 'book antiqua', palatino;">Use the answer to the exercise in the <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/1d9c959274674bda8930a7022cd47b33">preceding section</a> to calculate the values of \(f_{\text{right}}(f_{\text{up}}((0,1,0)))\) and \(f_{\text{up}}(f_{\text{right}}((0,1,0)))\). (Assume that \(\theta = \text{arccos}(1/3) \approx 70.53^\circ\).)</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Note:</i> This is a difficult exercise, unless you're comfortable with trigonometry and matrix multiplication. If you're not, you might find it helpful to look at the answer.</span>
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<h2 class="hd hd-2 unit-title">Duplicating the Graph</h2>
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<p><span style="font-family: book antiqua, palatino;">Let \(C\) be the set of Cayley Paths, and let \(C^e\) be the set of "endpoints'' of paths in \(C\). Since we are considering a version of the Cayley Graph that is wrapped around the surface of the ball, \(C^e\) consists of every point \(f_{\rho_1}(f_{\rho_2}(\dots f_{\rho_n}(c)\dots))\) for \(\langle\rho_1,\rho_2,\dots,\rho_n\rangle\) a Cayley Path.</span><br /><br /><span style="font-family: book antiqua, palatino;">In this section we will prove a preliminary version of our main result.</span></p>
<p><span style="font-family: book antiqua, palatino;">We will show that there is a decomposition of \(C^e\) into finitely many pieces which is such that the pieces can be rearranged -- without modifying their sizes or shapes -- so as to get two copies of \(C^e\). (The reason this is only a preliminary version of the main result is that \(C^e\) contains only countably many points from the ball's surface, and we're ultimately looking for a decomposition of the entire ball.)</span><br /><br /><span style="font-family: book antiqua, palatino;">Just as we did in in (<a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/2123814632114cd3855fad00c7b45d96" target="[object Object]">the abstract version of</a>) Warm-Up Case 3, we begin by considering the following four subsets of \(C\):</span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">\(U\) is the set of Cayley Paths that have ``up'' as their first step;</span></li>
<li><span style="font-family: book antiqua, palatino;">\(D\) is the set of Cayley Paths that have ``down'' as their first step;</span></li>
<li><span style="font-family: book antiqua, palatino;">\(L\) is the set of Cayley Paths that have ``left'' as their first step;</span></li>
<li><span style="font-family: book antiqua, palatino;">\(R\) is the set of Cayley Paths that have ``right'' as their first step;</span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">These sets constitute a partition of \(C\) (ignoring the empty path \(\langle\rangle\)). As before, we define \(\stackrel{\leftarrow}{X}\) as the result of deleting the first step from each Cayley Path in \(X\). And, as before, we get:</span></p>
<p style="padding-left: 30px;"><span style="font-family: book antiqua, palatino;">(\(\alpha\)) \(C = \stackrel{\leftarrow}{R} \cup \, L\)</span></p>
<p style="padding-left: 30px;"><span style="font-family: book antiqua, palatino;">(\(\beta\)) \(C = \stackrel{\leftarrow}{D} \cup \, U\)</span></p>
<p><span style="font-family: book antiqua, palatino;">which entail:</span></p>
<p style="padding-left: 30px;"><span style="font-family: book antiqua, palatino;">(\(\alpha'\)) \(C^e = \left(\stackrel{\leftarrow}{R}\right)^e \cup \, L^e\)</span><br /><br /><span style="font-family: book antiqua, palatino;">(\(\beta'\)) \(C^e = \left(\stackrel{\leftarrow}{D}\right)^e \cup \, U^e\)</span></p>
<p><span style="font-family: book antiqua, palatino;">Notice, moreover, that equations \((\alpha')\) and \((\beta')\) deliver our preliminary result, as long as we have the following:</span></p>
<ol>
<li><span style="font-family: book antiqua, palatino;">\(C^e\) is decomposed into \(U^e\), \(D^e\), \(L^e\) and \(R^e\) (ignoring the central vertex).</span></li>
<li><span style="font-family: book antiqua, palatino;">One can get from \(R^e\) to \(\left(\stackrel{\leftarrow}{R}\right)^e\), and from \(D^e\) to \(\left(\stackrel{\leftarrow}{D}\right)^e\), by performing a rotation.</span></li>
</ol>
<p><span style="font-family: book antiqua, palatino;">For when Claims 1 and 2 are in place, it follows from equations \((\alpha')\) and \((\beta')\) that \(C^e\) can be decomposed into \(U^e\), \(D^e\), \(L^e\) and \(R^e\) (ignoring the central point \(c\)) and recombined into two copies of \(C^e\) by performing suitable rotations.</span><br /><br /><span style="font-family: book antiqua, palatino;">In order to verify Claims 1 and 2, however, we'll need to proceed differently than in Warm-Up Case 3, because the details depend essentially on our interpretation of "following a Cayley Path''. Fortunately, it is all very straightforward:</span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">If you worked out the exercises in (<a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/2123814632114cd3855fad00c7b45d96" target="[object Object]">the abstract version of</a>) Warm-Up Case 3, you know that all it takes for claim 1 to be true is for different Cayley Paths to have different endpoints. And this is indeed the case: it is the Key Property of rotations, which we discussed <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/073583a9a6df44f3bdbfc90ef18ec867" target="[object Object]">earlier</a>.</span></li>
</ul>
<ul>
<li><span style="font-family: book antiqua, palatino;">The proof of claim 2 is the nicest feature of the entire construction.</span><br /><br /><span style="font-family: book antiqua, palatino;">Let \(\langle\rho_{\text{right}},\rho_2,\rho_3,\dots,\rho_n\rangle\) be a path in \(R\). Its endpoint is</span><br /><span style="font-family: book antiqua, palatino;">\[f_{{\text{right}}}(f_{\rho_2}(f_{\rho_3}(\dots f_{\rho_n}(c)\dots)))\] </span><br /><span style="font-family: book antiqua, palatino;">The corresponding path in \(\stackrel{\leftarrow}{R}\) is \(\langle\rho_2,\rho_3,\dots,\rho_n\rangle\) and its endpoint is:</span><br /><span style="font-family: book antiqua, palatino;">\[f_{\rho_2}(f_{\rho_3}(\dots f_{\rho_n}(c)\dots))\]</span><br /><span style="font-family: book antiqua, palatino;">So: <em>one can get from the endpoint of</em> \(\langle\rho_2,\rho_3,\dots,\rho_n\rangle\) <em>to the endpoint of</em> \(\langle\rho_{\text{right}},\rho_2,\rho_3,\dots,\rho_n\rangle\) <em>by applying</em> \(f_{\text{right}}\), <em>which is a "right" rotation!</em> </span><br /><br /><span style="font-family: book antiqua, palatino;">And, of course, the same is true for every other path in \(R\). So we get the result that one can move from the entire set \(\left(\stackrel{\leftarrow}{R}\right)^e\) to the entire set set \(R^e\) by performing a "right'' rotation (and therefore that one can move from \(R^e\) to \(\left(\stackrel{\leftarrow}{R}\right)^e\) by performing a "left'' rotation).</span><br /><br /><span style="font-family: book antiqua, palatino;">For analogous reasons, we get the result that one can move from \(\left(\stackrel{\leftarrow}{D}\right)^e\) to \(D^e\) by performing a "down'' rotation (and therefore that one can move from \(D^e\) to \(\left(\stackrel{\leftarrow}{D}\right)^e\) by performing an "up'' rotation).</span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">This concludes our preliminary result. We have verified that it is possible to decompose \(C^e\) into four separate components (\(U^e, D^e, L^e, R^e\)), and reassemble them to create two copies of \(C^e\), <em>without modifying sizes or shapes</em>. All we need to do is rotate \(R^e\) and \(D^e\). </span><br /><br /></p>
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<span style="font-family: 'book antiqua', palatino;">Recall that we have simplified things slightly by ignoring the "center" of the graph.</span>
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<span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(U, D, L, R\)</span> constitute a partition of <span class="math inline">\(C\)</span> (minus the empty path <span class="math inline">\(\langle \rangle\)</span>). Is it also true that <span class="math inline">\(U^e, D^e, L^e, R^e\)</span> constitute a partition of <span class="math inline">\(C^e\)</span> (minus the "center" point <span class="math inline">\(c\)</span>)?</span>
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<h2 class="hd hd-2 unit-title">Many Cayleys on a Ball</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">The next step is to duplicate the entire surface of our ball. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">The basic idea is straightforward. We have seen that there is a procedure that can be used duplicate the endpoints of a modified Cayley Graph that is wrapped around the surface of the ball. What we'll do now is work with an infinite family of modified Cayley Graphs, whose endpoints <em>jointly</em> cover the surface of the ball, without overlapping. We'll then duplicate the surface of the ball by applying our duplication procedure simultaneously to each of our infinitely many graphs.<br /><br />We begin by partitioning the surface of the sphere into cells. This is done by stipulating that two points are in the same cell if and only if one can get from one point to the other by applying a finite sequence of "left", "right", "up" and "down" rotations. <br /><br />Note that a cell is just the set of endpoints of some modified Cayley Graph, except that no endpoint has been designated as a "center''. So all one needs to do to characterize a modified Cayley Graph on the basis of a cell is to designate one of the members of the cell as a "center". </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Unfortunately, there turns out to be no way of specifying a condition that would succeed in singling out a unique member from an arbitrarily given cell. So our proof must make use of the Axiom of Choice. Rather than specifying a center for each cell, we use the Axiom of Choice to prove that a set of cell-centers exists, and therefore that there exists a family, \(\mathcal{G}\), of modified Cayley Graphs whose endpoints cover the surface of the sphere (without overlapping). <br /><br />Showing that \(\mathcal{G}\) must exist was the hard part. The remainder is straightforward. For each graph \(g\) in \(\mathcal{G}\), let \(C_g\) be the set of paths in \(g\). As usual, we decompose \(C_g\) (minus the empty path) into four non-overlapping pieces:</span></p>
<ul>
<li><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">\(U_g\) is the set of paths in \(g\) that have ``up'' as their first step;</span></li>
<li><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">\(D_g\) is the set of paths in \(g\) that have ``down'' as their first step;</span></li>
<li><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">\(L_g\) is the set of paths in \(g\) that have ``left'' as their first step;</span></li>
<li><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">\(R_g\) is the set of paths in \(g\) that have ``right'' as their first step.</span></li>
</ul>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Now, the upshot of the <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/a396432ef17d4b20b3d51b086f7f1874" target="[object Object]">preceding section</a> is that \(C_g^e\) (minus its "center") can be decomposed into pieces and that the pieces can be rearranged so as to get two copies of \(C_g^e\). We can get a first copy of \(C_g^e\) using \(L_g^e\) and \(R_g^e\) because<br />\[C_g^e = \left(\stackrel{\leftarrow}{R_g}\right)^e \cup \, L_g^e\]<br />and because one can move from \(R_g^e\) to \(\left(\stackrel{\leftarrow}{R}\right)^e\) by performing a "left" rotation. And we can get a second copy of \(C_g^e\) using \(D_g^e\) and \(U_g^e\) because \[C_g^e = \left(\stackrel{\leftarrow}{D_g}\right)^e \cup \, U_g^e\]<br />and because one can move from \(D_g^e\) to \(\left(\stackrel{\leftarrow}{D}\right)^e\) by performing an "up'' rotation.<br /><br />The fact that this is true for each modified Cayley Graph \(g\) in \(\mathcal{G}\) can be used to generate a version of the construction that applies <em>simultaneously</em> to every graph \(\mathcal{G}\). </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Let \(U_\mathcal{G}^e\) be the union of \(U^e_g\) for every \(g\) in \(\mathcal{G}\) (and similarly for \(D_\mathcal{G}^e\), \(L_\mathcal{G}^e\) and \(R_\mathcal{G}^e\)). </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">One can generate a first copy of the surface of the sphere (minus the "centers'' of graphs in \(\mathcal{G}\)) by combining \(L_\mathcal{G}^e\) plus a rotated version of \(R_\mathcal{G}^e\). And one can generate a second copy of the surface of the sphere (minus the "centers" of graphs in \(\mathcal{G}\)) by combining \(U_\mathcal{G}^e\) plus a rotated version of \(D_\mathcal{G}^e\). As usual, one can deal the "centers" by using the technique described in the answer to the exercise at the end of </span><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-family: book antiqua, palatino;"> (<a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/2123814632114cd3855fad00c7b45d96" target="[object Object]">the abstract version of</a>) Warm-Up Case 3,</span>. <br /><br />(I have been ignoring an important complication. As noted in earlier, it is essential to our procedure that different Cayley Paths have different endpoints. But, as noted <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/a396432ef17d4b20b3d51b086f7f1874" target="[object Object]">earlier</a>, we only get this Key Property for certain choices of \(c\). The reason is that, for each Cayley Path \(\langle\rho_1,\rho_2,\rho_3,\dots\rangle\), the rotation \(f_{\rho_1}(f_{\rho_2}(f_{\rho_r}(\dots)))\) does not change the location of the points intersecting its axis of rotation. So whenever \(c\) is such that some such point is the endpoint of some Cayley Path, we will have different Cayley Paths sharing an endpoint. Fortunately there are only countably many such points (because there are only countably many Cayley Paths) and therefore only countably many graphs in \(\mathcal{G}\) containing such points. One can deal with the vertices of these graphs separately, by applying a sophisticated version of </span><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">the trick we used in <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/96bae303e758441cb2f3a0a7bbd15423" target="[object Object]">Warm-Up Case 2</a> to turn a circle missing a point into a complete circle For details, check out the resources <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/6b0b5253672e4f13b3bbc679bcc898b7" target="[object Object]">here</a>.</span>)<br /><br />Success! We have managed to decompose the surface of our ball into finitely many pieces, and reassemble the pieces---without changing their sizes or shapes---so as to get two copies of the original surface. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">And now that we know how to duplicate the <em>surface</em> of our ball, it is easyto duplicate the ball itself. We can use the same procedure as before, but rather than working with points on the surface of the ball, we work with the lines that connect the center of the ball with each point.</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">That’s the Banach-Tarski Theorem! </span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">As a reward for your efforts, let me tell you a joke. ( I don’t know who came up with it originally; I heard about it from one of my students, Anthony Liu.) </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><em>Question:</em> “What’s an anagram of ‘Banach-Tarski’?” </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><em>Answer:</em> ”Banach-Tarski Banach-Tarski.”</span></p>
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<h3 class="hd hd-2">Video Review: Many Cayleys on a Melon</h3>
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<span style="font-family: 'book antiqua', palatino;">The procedure described above doesn't cover the center of the ball. Address this point by showing that one can start with a ball that is missing its central point, decompose it into two pieces, and reassemble the pieces (without changing their size or shape) so as to get a full ball.</span>
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<h2 class="hd hd-2 unit-title">A Demonstration</h2>
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