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<h2 class="hd hd-2 unit-title">The Banach-Tarski Theorem: Three Warm-Up Cases</h2>
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<h4><span style="font-family: 'book antiqua', palatino;">Three Warm-Up Cases</span></h4>
<p><span style="font-family: 'book antiqua', palatino;">Before tackling the Banach-Tarski Theorem, it’ll be useful to consider some warm-up cases. They are all constructions in which one decomposes an object into pieces and reassembles the pieces so as to end up with more than what we started with. But they are also simpler than the Banach-Tarski Theorem, and help illustrate ideas that will become important later.</span></p>
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<h2 class="hd hd-2 unit-title">Warm-up Case 1: A Line</h2>
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<p><span style="font-family: 'book antiqua', palatino;">Recall that <span class="math inline">\([0,\infty)\)</span> is the set of non-negative real numbers, and consider the result of removing the number 1 from that set: <span class="math inline">\([0,\infty) - \{1\}\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Is it possible to decompose <span class="math inline">\([0,\infty) - \{1\}\)</span> into two distinct parts, and reassemble the parts (without changing their size or shape) so as to get back <span class="math inline">\([0,\infty)\)</span>, including the number 1?</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Yes! In fact, it is very easy to do so. We’ll use a trick that is not too different from the trick we used to find room for Oscar in Hilbert’s Hotel, back in Chapter 1. Let us start by drawing picture of <span class="math inline">\([0,\infty) - \{1\}:\)</span></span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-16.png" type="saveimage" target="[object Object]" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now focus on a particular “piece” of <span class="math inline">\([0,\infty) - \{1\}\)</span>; namely, the set <span class="math inline">\(\{2,3,4,\ldots\}\)</span>. In the diagram below I have separated that “piece” from the rest of <span class="math inline">\([0,\infty) - \{1\}\)</span>:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-16final.png" type="saveimage" target="[object Object]" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now translate the “piece” one unit to the left:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-16final2.png" alt="" type="saveimage" target="[object Object]" preventdefault="function (){r.isDefaultPrevented=n}" stoppropagation="function (){r.isPropagationStopped=n}" stopimmediatepropagation="function (){r.isImmediatePropagationStopped=n}" isdefaultprevented="function t(){return!1}" ispropagationstopped="function t(){return!1}" isimmediatepropagationstopped="function t(){return!1}" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Finally, combine the translated “piece” with the rest of <span class="math inline">\([0,\infty) - \{1\}\)</span>:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-16final3.png" type="saveimage" target="[object Object]" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Success! We have decomposed the set <span class="math inline">\([0,\infty) - \{1\}\)</span> into two pieces, reassembled the pieces (without changing their size or shape), and ended up with the full set <span class="math inline">\([0,\infty)\)</span>.</span></p>
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<h3 class="hd hd-2">Video Review: Warm-Up Case 1</h3>
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<h2 class="hd hd-2 unit-title">Warm-Up Case 2: A Circle</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> In Warm-Up Case 1 we made essential use of the fact that the set [mathjaxinline][0,\infty) - \{1\}[/mathjaxinline] extends infinitely in one direction. We will now perform a similar decomposition involving an object of finite length. Let [mathjaxinline]S^1[/mathjaxinline] be a circle of radius 1, and consider the result of removing a single point [mathjaxinline]p[/mathjaxinline] from [mathjaxinline]S^1[/mathjaxinline]:</span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-17.png" height="50%" width="50%" type="saveimage" target="[object Object]" /></span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> Is it possible to decompose [mathjaxinline]S^1 - \{p\}[/mathjaxinline] into two distinct parts and reassemble the parts (without changing their size or shape) so as to get [mathjaxinline]S^1[/mathjaxinline]?</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> Yes! But this time we’ll have to be clever about how we decompose [mathjaxinline]S^1 - \{p\}[/mathjaxinline]. Let us characterize a subset [mathjaxinline]B[/mathjaxinline] of [mathjaxinline]S^1 - \{p\}[/mathjaxinline] by going clockwise around the circle. The first element of [mathjaxinline]B[/mathjaxinline] is the point [mathjaxinline]p_1[/mathjaxinline] which is 1 unit clockwise from [mathjaxinline]p[/mathjaxinline]. The second element of [mathjaxinline]B[/mathjaxinline] is the point [mathjaxinline]p_2[/mathjaxinline] which is 2 units clockwise from p. And so forth. Here are the locations of the first six elements of [mathjaxinline]B[/mathjaxinline]:</span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-17final.png" height="50%" width="50%" type="saveimage" target="[object Object]" /></span></p>
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<p><span style="font-family: 'book antiqua', palatino;">Since the radius of [mathjaxinline]S^1[/mathjaxinline] is one unit, its circumference is [mathjaxinline]2\pi[/mathjaxinline] units. As I’ll ask you to prove below, the fact that [mathjaxinline]2\pi[/mathjaxinline] is an irrational number can be used to prove that the points [mathjaxinline]p_1, p_2, p_3, \ldots[/mathjaxinline] are all distinct. In other words: if one starts at [mathjaxinline]p[/mathjaxinline] moves one unit clockwise, then another unit clockwise, then another, and so forth, one will never return to a location one had already visited.</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> Let us now <em>rotate</em> each point in [mathjaxinline]B[/mathjaxinline] one unit counter-clockwise. The result of such a rotation is the <em>unbroken</em> circle [mathjaxinline]S^1[/mathjaxinline], since [mathjaxinline]p_1[/mathjaxinline] ends up at [mathjaxinline]p[/mathjaxinline], [mathjaxinline]p_2[/mathjaxinline] ends up at [mathjaxinline]p_1[/mathjaxinline], [mathjaxinline]p_3[/mathjaxinline] ends up at [mathjaxinline]p_2[/mathjaxinline], and so forth. Success! We have decomposed our broken circle into two pieces, reassembled them (without changing their size of shape), and ended up with an unbroken circle!</span></p>
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<h3 class="hd hd-2">Video Review: Warm-Up Case 2</h3>
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<span style="font-family: 'book antiqua', palatino;">Show that the points <span class="math inline">\(p_1, p_2, p_3, \ldots\)</span> are all distinct.</span>
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<h2 class="hd hd-2 unit-title">Warm-Up Case 3: The Cayley Graph</h2>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> The point of the last two warm-up cases was to give you general flavor of how it is possible to decompose an object into pieces and reassemble the pieces so as to end up with more than one started with. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Now I’d like to you introduce you to a construction that is closely analogous to the construction we’ll use in proving the Banach-Tarski Theorem. It is also an example of dividing an object into finitely many pieces, reassembling the pieces, and getting two copies of the original object. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">There is, however, an important respect in which our example will be different than the Banach-Tarski Theorem. In our example, but not in the theorem, we’ll be allowed to change the size of the pieces as we move them around. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Consider the <strong>Cayley Graph,</strong> which is named after its creator, the British mathematician Arthur Cayley:</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-18.png" type="saveimage" target="[object Object]" width="687" height="629" /></span></p>
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<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">The best way to think of a Cayley Graph is as an infinite collection of “Cayley Paths”. A Cayley Path is a finite sequence of steps, starting from a central point, [mathjaxinline]c[/mathjaxinline]. Each step is taken in one of four directions: up, down, right or left, with the important restriction that one is forbidden from following opposite directions in adjacent steps. (One cannot, for example, follow an “up” step with a “down” step, or a “left” step with a “right” step.) </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">The steps in a Cayley Path span smaller and smaller distances: in the first step of a Cayley Path you advance one unit; in the second step you advance half a unit; in the third step you advance a quarter of a unit; and so forth. </span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">An <strong>endpoint</strong> (or vertex) of the Cayley Graph is simply the endpoint of some Cayley Path. (We count the “empty” path as a Cayley Path, and therefore count the starting point c as an endpoint of the graph.)</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> Let [mathjaxinline]C[/mathjaxinline] be the set of every path in the Cayley Graph, and let [mathjaxinline]C^e[/mathjaxinline] be the set of endpoints in [mathjaxinline]C[/mathjaxinline]. (In general, if \(X\) is a set of Cayley Paths, we will let \(X^e\) be the set of endpoints of paths in \(X\).)</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">We will see that it is possible to divide [mathjaxinline]C^e[/mathjaxinline] into a finite number of components, rearrange the components and end up with two copies of [mathjaxinline]C^e[/mathjaxinline]. We will allow some of our pieces to change size as we move them around here, but remember that this won’t be allowed when we prove the Banach-Tarski Theorem.</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"> The first step of our construction is to partition the set of Cayley Paths into four cells: </span></p>
<ul>
<li><span style="font-family: 'book antiqua', palatino;">[mathjaxinline]U[/mathjaxinline] is the set of Cayley Paths that have “up” as their first step;</span></li>
<li><span style="font-family: 'book antiqua', palatino;">[mathjaxinline]D[/mathjaxinline] is the set of Cayley Paths that have “down” as their first step;</span></li>
<li><span style="font-family: 'book antiqua', palatino;">[mathjaxinline]L[/mathjaxinline] is the set of Cayley Paths that have “left” as their first step;</span></li>
<li><span style="font-family: 'book antiqua', palatino;">[mathjaxinline]R[/mathjaxinline] is the set of Cayley Paths that have “right” as their first step;</span></li>
</ul>
<p><span style="font-family: 'book antiqua', palatino;">(I am simplifying things slightly by ignoring the empty path [mathjaxinline]\langle\rangle[/mathjaxinline]. You’ll be asked to find a version of the construction that does not require this simplification in the exercise below.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Think of each of our four sets as a different “quadrant” of our Cayley Graph. In the diagram below, I’ve used a different color for each. (Since we’re ignoring the central point [mathjaxinline]c[/mathjaxinline] for now, I’ve left it out of the diagram.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-19.png" type="saveimage" target="[object Object]" width="794" height="831" /></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let [mathjaxinline]T^e, B^e, L^e [/mathjaxinline] and [mathjaxinline]R^e[/mathjaxinline] be the sets of endpoints of [mathjaxinline]T, B, L[/mathjaxinline] and [mathjaxinline]R[/mathjaxinline], respectively. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">If you focus your attention on [mathjaxinline]R^e[/mathjaxinline], you’ll see that it is an exact copy of [mathjaxinline]C^e - L^e[/mathjaxinline], except that it is a bit shrunken in size, and translated slightly to the right. This means that we can use [mathjaxinline]R^e[/mathjaxinline] and [mathjaxinline]L^e[/mathjaxinline] to create a perfect copy of [mathjaxinline]C^e[/mathjaxinline]: we start by expanding [mathjaxinline]R^e[/mathjaxinline], and then translate it slightly to the left to meet [mathjaxinline]L^e[/mathjaxinline], as in the following diagram. (I’ve included Cayley Paths along with their endpoints to make the diagram easier to read.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><img src="https://studio.edx.org/asset-v1:MITx+24.118x+1T2018+type@asset+block@FiguresDiagramsParadoxInfinity-20.png" type="saveimage" target="[object Object]" width="730" height="805" /></span></p>
<p></p>
<p><span style="font-family: 'book antiqua', palatino;">Similarly, we can use [mathjaxinline]T^e[/mathjaxinline] and [mathjaxinline]B^e[/mathjaxinline] to create a perfect copy of [mathjaxinline]C^e[/mathjaxinline]: we start by expanding [mathjaxinline]D^e[/mathjaxinline], and then translate it slightly upward to meet [mathjaxinline]T^e[/mathjaxinline].</span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;">Here's an animation of how this would work. </span><span style="font-family: book antiqua,palatino;">(Animation, courtesy of </span><span lang="en-US"><span style="font-family: Times New Roman,serif; font-size: medium;" size="3" face="Times New Roman,serif"><span style="font-size: 12pt;"><span style="font-family: book antiqua,palatino;"><span style="color: #1f497d; font-size: small;" color="#1F497D" size="2" face="Calibri,sans-serif"><span style="font-size: 11pt;"></span></span>קובי כרמל on Wikimedia. License CC BY-SA. See <a href="http://commons.wikimedia.org/wiki/File:Cayley_backward.gif" target="[object Object]">here</a></span><span style="font-family: book antiqua,palatino;">.)</span></span></span></span></p>
<p><span style="font-size: 12pt; font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><img src="http://upload.wikimedia.org/wikipedia/commons/0/03/Cayley_backward.gif" alt="Cayley copy" type="saveimage" target="[object Object]" preventdefault="function (){r.isDefaultPrevented=n}" stoppropagation="function (){r.isPropagationStopped=n}" stopimmediatepropagation="function (){r.isImmediatePropagationStopped=n}" isdefaultprevented="function t(){return!1}" ispropagationstopped="function t(){return!1}" isimmediatepropagationstopped="function t(){return!1}" width="750" height="563" /></span></span></p>
<p><span style="font-family: book antiqua,palatino;"><br /></span><span lang="en-US"></span><span style="font-family: 'book antiqua', palatino;">Success! We have decompose [mathjaxinline]C^e[/mathjaxinline] into finitely many pieces, reassembled them (changing sizes but not shapes), and ended up with two copies of [mathjaxinline]C^e[/mathjaxinline]!</span></p>
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<span style="font-family: 'book antiqua', palatino;">I simplified the construction above by ignoring the empty path <span class="math inline">\(\langle \rangle\)</span>, whose endpoint is the graph&#8217;s central vertex. Find a version of the construction that does not require this simplification.</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Hint:</i> Proceed by dividing <span class="math inline">\(C^e\)</span> into the following components: <span class="math inline">\(U^e, D^e, (L\cup S)^e, (R-S)^e\)</span>, where <span class="math inline">\(S\)</span> is the set of Cayley Paths that contain only &#8220;right&#8221; steps: <span class="math display">\[S = \{\langle \rangle, \langle\text{right}\rangle, \langle\text{right, right}\rangle, \langle\text{right, right, right}\rangle, \dots\}\]</span> As usual <span class="math inline">\(X^e\)</span> is the set of endpoints of Cayley Paths in <span class="math inline">\(X\)</span>.</span>
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<h2 class="hd hd-2 unit-title">A more abstract description of the procedure</h2>
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<p><span style="font-family: book antiqua, palatino;">In this section we will replicate the procedure of the preceding section in a more abstract setting. It is worth doing so because the more abstract version of the procedure is exactly analogous to the procedure we'll use to prove the Banach-Tarski Theorem.</span></p>
<p><span style="font-family: book antiqua, palatino;">First, some notation: if \(X\) is a set of Cayley Paths, let \(\stackrel{\leftarrow}{X}\) be the set that results from eliminating the first step from each of the Cayley Paths in \(X\). </span></p>
<p><span style="font-family: book antiqua, palatino;">Now recall that \(R\) is the set of Cayley Paths that start with a "right'' step. Since one can get a valid Cayley path by appending a "right'' step to the beginning of any Cayley path that does not start with a "left'' step, \(\stackrel{\leftarrow}{R} \, = C - L\). (For analogous reasons, \(\stackrel{\leftarrow}{D} \, = C - U\).) So we have:</span></p>
<p><span style="font-family: book antiqua, palatino;">(\(\alpha\)) \(C = \stackrel{\leftarrow}{R} \cup \, L\)</span></p>
<p><span style="font-family: book antiqua, palatino;">(\(\beta\)) \(C = \stackrel{\leftarrow}{D} \cup \, U\)</span><br /><br /><span style="font-family: book antiqua, palatino;">But since every Cayley Path has a unique endpoint, equations \((\alpha)\) and \((\beta)\) immediately entail:</span></p>
<p><span style="font-family: book antiqua, palatino;">(\(\alpha'\)) \(C^e = \left(\stackrel{\leftarrow}{R}\right)^e \cup \, L^e\)</span></p>
<p><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, however, that equations \((\alpha')\) and \((\beta')\) deliver the result of the previous section. This is because (as you'll be asked to verify below) we have have each of 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 translation together with an expansion. </span></li>
</ol>
<p><span style="font-family: book antiqua, palatino;">So 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 vertex), and recombined into two copies of \(C^c\) by carrying out suitable translations and expansions.</span><br /><br /><span style="font-family: book antiqua, palatino;">This procedure is exactly analogous to the procedure we'll use to prove the Banach-Tarski Theorem. The only difference is that we'll work with a variant of the Cayley Graph in which "following a Cayley Path'' has a different geometrical interpretation than it does here. We'll think of our Cayley Graph as wrapped around the surface of a ball, and we'll think of each step of a Cayley Path as corresponding to a certain <em>rotation</em> of the ball.</span><br /><br /><span style="font-family: book antiqua, palatino;">The key reason we'll be able to prove our result in the new setting is that the truth of equations \((\alpha')\) and \((\beta')\) is independent of how we interpret "following a Cayley Path''. </span></p>
<p><span style="font-family: book antiqua, palatino;">We'll need to proceed carefully, though, because the truth of claims 1 and 2 above does depend on how we interpret "following a Cayley Path''. Fortunately, this won't be a problem: </span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">We'll be able to recover claim 1 because our interpretation of "following a Cayley Path'' will be such that different Cayley Paths always have different endpoints. </span></li>
<li><span style="font-family: book antiqua, palatino;">We won't be able to recover claim 2 as it is stated above, but we'll get something even better: our new interpretation of "following a Cayley Path'' will yield the result that 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 suitable rotation -- no expansions required. This means that we'll be able to reassemble \(U^e\), \(D^e\), \(L^e\) and \(R^e\) into two copies of \(C^e\) <em>without changing their size</em>.</span></li>
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<span style="font-family: 'book antiqua', palatino;">Verify that \(C^e\) is decomposed into \(U^e\), \(D^e\), \(L^e\) and \(R^e\). (Ignore the central vertex.)</span>
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<span style="font-family: 'book antiqua', palatino;">Verify that 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 translation together with an expansion.</span>
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Problem 3
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<span style="font-family: 'book antiqua', palatino;">As before, I have simplified the construction by ignoring the empty path, whose endpoint is the center of the Cayley Graph. The exercise at the end of the previous section shows that it is possible to revise the construction of the Caley Graph in such a way that this simplification is not required. Show that the same trick can be used in the more abstract setting too.</span>
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<span style="font-family: 'book antiqua', palatino;"><i>Hint:</i> Proceed by dividing \(C^e\) into the following components: \(U^e, D^e, (L\cup S)^e, (R-S)^e\), where \(S\) is the set of Cayley Paths that contain only "right'' steps:
\[S = \{\langle\rangle, \langle\text{right}\rangle, \langle\text{right, right}\rangle, \langle\text{right, right, right}\rangle, \dots\}\]
As usual \(X^e\) is the set of endpoints of Cayley Paths in \(X\).</span>
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