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<h2 class="hd hd-2 unit-title">Ordinal Addition</h2>
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<p class="p1"><span style="font-family: 'book antiqua', palatino;"><strong><span style="font-size: 12pt;">Ordinal Addition</span></strong></span></p>
<p><span style="font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">A nice feature of the hierarchy of ordinals is that it allows for arithmetical operations. </span><span style="font-size: 12pt;">Let me start by telling you how to define ordinal addition. I'll then tell you how to define ordinal multiplication. <br /></span></span></p>
<p><span style="font-family: 'book antiqua', palatino;"></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here is the intuitive idea behind ordinal addition. A well-ordering of type <span class="math inline">\((\alpha+\beta)\)</span> is the result of starting with a well-ordering of type <span class="math inline">\(\alpha\)</span> and appending a well-ordering of type <span class="math inline">\(\beta\)</span> at the end.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Consider, for example, the ordinal <span class="math inline">\((\omega+0')\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">What well-order type does it represent? It is the well-order type that one gets to by starting with a well-ordering of type <span class="math inline">\(\omega\)</span> (i.e. "<span class="math inline">\(| | | | \dots\)"</span>) and appending an ordering of type <span class="math inline">\(0'\)</span> (i.e. "<span class="math inline">\(|\)"</span>) at the end: <span class="math display">\[| | | | \dots |\]</span> So <span class="math inline">\((\omega+0')\)</span> represents the well-order type "<span class="math inline">\(| | | | \dots |\)"</span>. And since <span class="math inline">\(\omega' = \{0,0',0'',\dots,\omega\}\)</span>, which is of type "<span class="math inline">\(| | | | \dots |\)"</span>, this means that <span class="math inline">\(\omega + 0'\)</span> is just <span class="math inline">\(\omega'\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now consider the ordinal <span class="math inline">\((\omega+\omega)\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">It is the result of appending a well-ordering of type <span class="math inline">\(\omega\)</span> (i.e. "<span class="math inline">\(| | | | \dots\)"</span>) to the end of itself: <span class="math display">\[| | | | \dots | | | | \dots\]</span> So <span class="math inline">\((\omega+\omega)\)</span> represents the well-order type "<span class="math inline">\(| | | | \dots | | | \dots\)"</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">An important difference between ordinal addition and regular addition is that ordinal addition is not <strong>commutative</strong>. In other words, it is not generally the case that <span class="math inline">\(\alpha + \beta = \beta + \alpha\)</span>. Notice, for example, that <span class="math inline">\((0'+\omega)\neq(\omega+0')\)</span>. For whereas <span class="math inline">\((\omega+0')\)</span> represents well-orderings of type "<span class="math inline">\(| | | | \dots |\)"</span>, <span class="math inline">\((0'+\omega)\)</span> represents well-orderings of type "<span class="math inline">\(| | | | \dots\)"</span>. So <span class="math inline">\((0'+\omega)=\omega\)</span>.</span></p>
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<span style="font-family: 'book antiqua', palatino;">Even though ordinal addition is not generally commutative, it does have the following property:</span>
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<strong>Associativity</strong>
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<br/>
<span style="font-family: 'book antiqua', palatino;">For any ordinals <span class="math inline">\(\alpha\)</span>, <span class="math inline">\(\beta\)</span> and <span class="math inline">\(\gamma\)</span>, <span class="math inline">\((\alpha+\beta)+\gamma = \alpha+(\beta+\gamma)\)</span></span>
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<span style="font-family: 'book antiqua', palatino;">Does that mean that the following is true: <span class="math inline">\((\omega+\omega)+\omega = \omega +(\omega +\omega )\)</span>? (If so, verify Associativity in that special case.)</span>
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<span style="font-family: 'book antiqua', palatino;">Yes, \((\omega+\omega)+\omega = \omega +(\omega +\omega )\) is a special case of Associativity, and I've done my best to prove it.</span>
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<span style="font-family: 'book antiqua', palatino;">No, \((\omega+\omega)+\omega = \omega +(\omega +\omega )\) is not a special case of Associativity.</span>
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<span style="font-family: 'book antiqua', palatino;">Although ordinal addition is not commutative in general, it is commutative in some cases.</span>
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<span style="font-family: 'book antiqua', palatino;">Is ordinal addition commutative in the following special case: <span class="math inline">\(0'+0''' = 0'''+0'\)</span>?</span>
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<span style="font-family: 'book antiqua', palatino;">Determine whether the following is true:<span class="math inline">\((\omega + 0') + (\omega + \omega) = (\omega + \omega) + (\omega + 0')\)</span>.</span>
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<h2 class="hd hd-2 unit-title">Ordinal Multiplication</h2>
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<p><span style="font-family: 'book antiqua', palatino;">Let us now turn to ordinal multiplication. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Here the intuitive idea is that a well-ordering of type <span class="math inline">\((\alpha\times\beta)\)</span> is the result of starting with a well-ordering of type <span class="math inline">\(\beta\)</span> and replacing each position in the ordering with a well-ordering of type <span class="math inline">\(\alpha\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Consider, for example, the ordinal <span class="math inline">\((\omega\times0'')\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">What well-order type does it represent? It is the well-order type that one gets by starting with a well-ordering of type <span class="math inline">\(0''\)</span> (i.e. "<span class="math inline">\( | | \)"</span>) and replacing each position with a well-ordering of type <span class="math inline">\(\omega\)</span> (i.e. "<span class="math inline">\( | | | | | \dots \)"</span>). Schematically, <span class="math display">\[\begin{array}{cc} | &|\\ \downarrow &\downarrow \\ \overbrace{| | | | | \dots} & \overbrace{| | | | | \dots} \end{array}\]</span> Since <span class="math inline">\((\omega+\omega)\)</span> is of type "<span class="math inline">\(| | | | | \dots | | | | | \dots\)"</span>, this means that <span class="math inline">\((\omega\times0'')=(\omega+\omega)\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">As in the case of ordinal addition, ordinal multiplication is not <strong>commutative</strong>: it is not generally the case that <span class="math inline">\(\alpha \times \beta = \beta \times \alpha\)</span>. Notice, for example, that <span class="math inline">\((\omega\times0'')\neq(0''\times\omega)\)</span>. For whereas we have seen that <span class="math inline">\((\omega\times0'')=(\omega+\omega)\)</span>, it is not the case that <span class="math inline">\((0''\times\omega)=(\omega+\omega)\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">To see this, recall that <span class="math inline">\((0''\times\omega)\)</span> is the result of starting with a well-ordering of type <span class="math inline">\(\omega\)</span> (i.e. "<span class="math inline">\(| | | | | \dots\)"</span>) and replacing each position in that ordering with a well-ordering of type <span class="math inline">\(0''\)</span> (i.e. "<span class="math inline">\(| |\)"</span>). Schematically: <span class="math display">\[\begin{array}{cccl} | & | & | & \dots\\ \downarrow & \downarrow & \downarrow \\ \overbrace{| |} &\overbrace{| |} & \overbrace{| |} &\dots \end{array}\]</span> The result is just a well-ordering of type "<span class="math inline">\(| | | | | \dots\)"</span>. So <span class="math inline">\((0''\times\omega)\)</span> is <span class="math inline">\(\omega\)</span>, rather than <span class="math inline">\(\omega+\omega\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let me mention a final example of ordinal multiplication. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Consider the ordinal <span class="math inline">\(\omega\times\omega\)</span>. What order type does it represent? It is the well-order type that one gets to by starting with a well-ordering of type <span class="math inline">\(\omega\)</span> (i.e. "<span class="math inline">\(| | | | | \dots\)"</span>) and replacing each position "<span class="math inline">\(|\)"</span> with an ordering of type <span class="math inline">\(\omega\)</span>: <span class="math display">\[\begin{array}{cccl} | & | & | & \dots\\ \downarrow & \downarrow & \downarrow \\ \overbrace{| | | \dots} &\overbrace{| | | \dots} & \overbrace{| | | \dots} & \text{\(\dots\)} \end{array}\]</span> So <span class="math inline">\(\omega\times\omega\)</span> is an ordering of type: <span class="math display">\[\underbrace{| | | | | \ldots | | | | | \ldots | | | | | \ldots \ \text{\(\dots\)}}_{\text{\(\omega\)-many times}}\]</span></span></p>
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<span style="font-family: 'book antiqua', palatino;">As in the case of ordinal addition, ordinal multiplication has the following property:</span>
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<strong>Associativity</strong>
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<span style="font-family: 'book antiqua', palatino;">For any ordinals <span class="math inline">\(\alpha\)</span>, <span class="math inline">\(\beta\)</span> and <span class="math inline">\(\gamma\)</span>, <span class="math inline">\((\alpha\times\beta)\times\gamma = \alpha\times(\beta\times\gamma)\)</span></span>
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<span style="font-family: 'book antiqua', palatino;">Does this mean that the following is true: <span class="math inline">\((\omega\times\omega)\times\omega=\omega\times(\omega\times\omega)\)</span>? (If so, verify Associativity in that special case.)</span>
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<span style="font-family: 'book antiqua', palatino;">Yes, \((\omega\times\omega)\times\omega=\omega\times(\omega\times\omega)\) is a special case of Associativity, and I've done my best to prove it.</span>
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<span style="font-family: 'book antiqua', palatino;">No, \((\omega\times\omega)\times\omega=\omega\times(\omega\times\omega)\) is not a special case of Associativity.</span>
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<span style="font-family: 'book antiqua', palatino;">Although ordinal multiplication is not commutative in general, it is commutative in some cases.</span>
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<span style="font-family: 'book antiqua', palatino;">Is ordinal multiplication commutative in the following special case: <span class="math inline">\((0''\times 0''') = (0'''\times 0'')\)</span>?</span>
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<span style="font-family: 'book antiqua', palatino;">Yes, it is.</span>
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<span style="font-family: 'book antiqua', palatino;">Unlike regular addition and multiplication, ordinal addition and multiplication fail to be <strong>distributive</strong>: it is not generally the case that for all ordinals <span class="math inline">\(\alpha\)</span>, <span class="math inline">\(\beta\)</span>, <span class="math inline">\(\gamma\)</span>, <span class="math inline">\((\alpha+\beta)\times \gamma = \alpha\times \gamma + \beta\times \gamma\)</span>.</span>
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<span style="font-family: 'book antiqua', palatino;">Is ordinal multiplication distributive in the following special case: <span class="math inline">\((\omega +0''')\times 0'' = (\omega\times 0'')+(0''' \times 0'')\)</span>?</span>
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<span style="font-family: 'book antiqua', palatino;">Ordinals are, however, <strong>distributive on the left</strong>. In other words: for all ordinals <span class="math inline">\(\alpha\)</span>, <span class="math inline">\(\beta\)</span>, <span class="math inline">\(\gamma\)</span>, <span class="math inline">\(\alpha\times (\beta+\gamma) = \alpha\times \beta + \alpha\times \gamma\)</span>.</span>
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<span style="font-family: 'book antiqua', palatino;">Verify that this is true in the following special case: <span class="math inline">\(0'' \times (\omega +0''') = (0'' \times \omega) + (0'' \times 0''') \)</span>.</span>
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<h2 class="hd hd-2 unit-title">Optional: A More Rigorous Characterization of Ordinal Addition and Multiplication</h2>
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<p><span style="font-family: 'book antiqua', palatino;">In the preceding subsections I gave you an informal characterization of ordinal addition and multiplication. Fortunately, it is easy to characterize these operations more rigorously: <span class="math display">\[\begin{array}{rclcl} \alpha &+ &0 &= &\alpha \\ \alpha &+ &\beta' &= &(\alpha + \beta)'\\ \alpha &+ &\lambda &= & \bigcup \{\alpha + \beta : \beta < \lambda\} \text{ ($\lambda$ a limit ordinal)}\\ \\ \alpha &\times &0 &= &0 \\ \alpha &\times &\beta' &= &(\alpha \times \beta) + \alpha\\ \alpha &\times &\lambda &= & \bigcup \{\alpha \times \beta : \beta < \lambda\} \text{ ($\lambda$ a limit ordinal)} \end{array}\]</span> Notice, incidentally, that there is no reason to stop at multiplication. We could use a similar strategy to define exponentiation: <span class="math display">\[\begin{array}{lcl} \alpha^0 &= & 0' \\ \alpha^{\beta'} &= &(\alpha^\beta) \times \alpha \\ \alpha^{\lambda} &= & \bigcup \{\alpha^\beta : \beta < \lambda\} \text{ ($\lambda$ a limit ordinal)} \end{array}\]</span> And we could go further still, defining ordinal <a href="https://en.wikipedia.org/wiki/Tetration" target="[object Object]">tetration</a>, and beyond.</span></p>
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<h3 class="hd hd-2">Ordinal Multiplication</h3>
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<p><span style="font-family: 'book antiqua', palatino;"><strong>Correction! </strong>The third formula for ordinal multiplication on the right of the screen is incorrect. The \(+\) should, of course, be \(\times\). So in full the formula should be:</span></p>
<p style="text-align: center;"><span style="font-family: 'book antiqua', palatino;">\( \alpha \times \lambda = \bigcup \{\alpha \times \beta : \beta < \lambda\} \)</span></p>
<p style="text-align: left;"><span style="font-family: 'book antiqua', palatino;">The narration and calculations are all correct.</span></p>
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<h5><span style="font-family: 'book antiqua', palatino;">Circularity</span></h5>
<p><span style="font-family: 'book antiqua', palatino;">You may have noticed that the definitions above are all <em>circular</em>. The definition of ordinal addition, for example, uses ordinal addition to define ordinal addition. But it is important to observe that the definitions are not circular in a problematic way: they all succeed in fixing the meanings of the operations they are meant to define.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Let me explain how this works by focusing on the case of addition. The basic insight is that although our definition of <span class="math inline">\(\alpha + \beta\)</span> might presuppose that certain other instances ordinal addition have been defined, the relevant instances will always be of the form of <span class="math inline">\(\alpha + \gamma\)</span> where <span class="math inline">\(\gamma <_o \beta\)</span>. It is useful to work through some examples. Notice, first, that <span class="math inline">\(\omega + 0\)</span> is well defined, since the first clause of our definition tells us that:</span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(\omega + 0 = \omega\)</span>.</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">Result (0) can then be used to define <span class="math inline">\(\omega + 0'\)</span>. For the second clause of our definition tells us that <span class="math inline">\(\omega + 0' = (\omega + 0)'\)</span>. So result <span class="math inline">\((0)\)</span> yields:</span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(\omega + 0' = \omega'\)</span>.</span></p>
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</ul>
<p><span style="font-family: 'book antiqua', palatino;">Result (1) can then be used to define <span class="math inline">\(\omega + 0''\)</span>. For the second clause of our definition tells us that <span class="math inline">\(\omega + 0'' = (\omega + 0')'\)</span>. So result (1) yields:</span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(\omega + 0'' = \omega''\)</span>.</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">This technique can be iterated to prove the following for each natural number <span class="math inline">\(n\)</span>:</span></p>
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<p><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\omega + 0\!\!\overbrace{'{^{\dots}}'}^{\text{$n$ times}} = \omega\!\!\overbrace{'{^{\dots}}'}^{\text{$n$ times}}\)</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">Results (0), (1), …can then be used together to define <span class="math inline">\(\omega + \omega\)</span>. Since <span class="math inline">\(\omega\)</span> is a limit ordinal, the third clause of our definition tells us that:</span></p>
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<p><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\omega + \omega = \bigcup \{\omega + \beta : \beta < \omega\}\)</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">But <span class="math inline">\( \bigcup \{\omega + \beta : \beta < \omega\}\)</span> is just <span class="math inline">\(\bigcup \{\omega + 0, \omega + 0', \omega + 0'',\dots\}\)</span>. By the definition of <span class="math inline">\(\bigcup\)</span>, this means that <span class="math inline">\(\omega + \omega\)</span> consists of the members of <span class="math inline">\(\omega + 0\!\!\overbrace{'{^{\dots}}'}^{\text{$n$ times}}\)</span>, for each <span class="math inline">\(n\)</span>. Since an ordinal is just the set of its predecessors, this gives us:</span></p>
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<li>
<p><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\omega + \omega = \{0,0',\dots,\omega, \omega', \omega'', \dots\}\)</span></p>
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