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<h2 class="hd hd-2 unit-title">An Informal Introduction to the Theorem</h2>
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<p><span style="color: #3c3c3c; font-family: 'book antiqua', palatino; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">We have seen that there are just as many natural numbers as rational numbers. Is this because there is a bijection between any two infinite sets? Amazingly, the answer is ‘no’. Some infinities are bigger than others! In this section, we will prove that there are more real numbers than there are natural numbers.</span></p>
<p><span style="color: #3c3c3c; font-family: 'book antiqua', palatino; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">If you’d like to start out with an informal introduction to the material, check out the video below. It was produced by Damien Rochford and Gaurav Vazirani as part of<span> </span></span><a href="http://wi-phi.com" target="[object Object]" style="color: #1d9dd9; font-family: 'book antiqua', palatino; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff;">Wi-Phi</a><span style="color: #3c3c3c; font-family: 'book antiqua', palatino; font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: #ffffff; text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">, a project which works with the Khan Academy to make philosophy more accessible to non-philosophers.</span></p>
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<h2 class="hd hd-2 unit-title">The Real Numbers: Preliminaries</h2>
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<p><span style="font-family: 'book antiqua', palatino;">We will now verify an astounding claim I made earlier: <em>some infinities are bigger than others.</em> We will show, in particular, that there are more real numbers than there are natural numbers.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The set of <strong>real numbers</strong>, <span class="math inline">\(\mathbb{R}\)</span>, is the set of numbers that you can write down in decimal notation by starting with a numeral (e.g. “17”), adding a decimal point (“17.”), and then an infinite sequence of digits (e.g. “17.8423…”). Any symbol between “0” and “9” is an admissible digit.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Every natural number is a real number, since every natural can all be written out as a numeral followed by an infinite sequence of zeroes (e.g. 17=17.0000…). </span></p>
<p><span style="font-family: 'book antiqua', palatino;">The rational numbers are also real numbers. In fact, they are real numbers with the special feature of having <em>periodic</em> decimal expansions: after a certain point, the expansions are just repetitions of some finite sequence of digits. For instance, <span class="math inline">\(1318/185 = 7.12(432)\)</span>, where the brackets surrounding “432” indicate that these three digits are to be repeated indefinitely, so as to get <span class="math inline">\(7.12432432432432\dots\)</span>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Not every real number is a rational number, however. There are irrational real numbers, such as <span class="math inline">\(\pi\)</span> and <span class="math inline">\(\sqrt 2\)</span>. Unlike the decimal expansions of rational numbers, the decimal expansions of irrational numbers are never periodic: they never end with an infinitely repeating a finite sequence of digits.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Some real numbers have multiple names in decimal notation. For example, the number \(1/2\) is named both by the numeral “<span class="math inline">\(0.5(0)\)</span>” and by the numeral “<span class="math inline">\(0.4(9)\)</span>”. To see this, note that each of the following inequalities must hold: <span class="math display">\[\begin{aligned} 0.5(0) - 0.4(9)& < &0.01 \text{ (since $0.49 < 0.4(9)$)}\\ 0.5(0) - 0.4(9)& < &0.001 \text{ (since $0.499 < 0.4(9)$)}\\ 0.5(0) - 0.4(9)& < &0.0001 \text{ (since $0.4999 < 0.4(9)$)}\\ &\vdots& \end{aligned}\]</span> This means that the difference between 0.5(0) and 0.4(9) must be smaller than each of <span class="math inline">\(0.01, 0.001, 0.0001\)</span>, and so forth. Since the difference between real numbers must be a non-negative real number, and since the only non-negative real number smaller than each of <span class="math inline">\(0.01, 0.001, 0.0001, \dots\)</span> is 0, it follows that the difference between 0.5(0) and 0.4(9) must be <span class="math inline">\(0\)</span>. So <span class="math inline">\(0.5(0) = 0.4(9)\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Having real numbers with multiple names in decimal notation turns out to be a nuisance in the present context. Fortunately, the problem is easily avoided. The crucial observation is that the only real numbers with multiple names are those named by numerals that end in an infinite sequence of 9s. Such numbers always have exactly two names: one ending in an infinite sequence of 9s and one ending in an infinite sequence of 0s. So we can avoid having numbers with multiple names by treating numerals that end in an infinite sequence of 9s as invalid, and naming the relevant numbers using numerals that end in an infinite sequence of 0s.</span></p>
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<span style="font-family: 'book antiqua', palatino;">In the text above I suggested avoiding the problem of multiple numerals by ignoring numerals that end in an infinite sequence of 9s. Could we have also avoided the problem by ignoring numerals that end in an infinite sequence of 0s?</span>
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<span style="font-family: 'book antiqua', palatino;">Yes. We could have also avoided the problem this way.</span>
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<span style="font-family: 'book antiqua', palatino;">No. We could not have avoided the problem this way.</span>
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<h3 class="hd hd-3 problem-header" id="5fd48ac7ba254d52987cdf6290c3bb1c-problem-title" aria-describedby="block-v1:MITx+24.118x+2T2020+type@problem+block@5fd48ac7ba254d52987cdf6290c3bb1c-problem-progress" tabindex="-1">
Problem 2
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>
<span style="font-family: 'book antiqua', palatino;">Distinct decimal expansions <span class="math inline">\(0.\delta^1_1\delta^1_2\delta^1_3\dots\)</span> and <span class="math inline">\(0.\delta^2_1\delta^2_2\delta^2_3\dots\)</span> can only refer to the same number if one of them is of the form \(0.s(9)\) and the other is of the form \(0.s'(0)\), where \(s\) is a sequence of digits and \(s'\) is the result of raising the last digit in s by one.</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">True or False?</span>
</p>
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<p class="answer" id="answer_5fd48ac7ba254d52987cdf6290c3bb1c_2_1"/>
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<span style="font-family: 'book antiqua', palatino;">(If it's true, try proving it. If it's false, find a counterexample.)</span>
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Problem 3
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<span style="font-family: 'book antiqua', palatino;">Every rational number has a periodic decimal expansion.</span>
</p>
<p>
<span style="font-family: 'book antiqua', palatino;">True or False?</span>
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<p class="answer" id="answer_0090c173dd55404d8cc304099b4c84bb_2_1"/>
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<span style="font-family: 'book antiqua', palatino;">(If it's true, try proving it. If it's false, find a counterexample.)</span>
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Problem 4
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<span style="font-family: 'book antiqua', palatino;">If \(p\) is a period of length \(n\), then \(0.(p) = \) ?</span>
</p>
<div class="choicegroup capa_inputtype" id="inputtype_a7dfb8f0181041fcac75b47ffa647167_2_1">
<fieldset aria-describedby="status_a7dfb8f0181041fcac75b47ffa647167_2_1">
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<input type="radio" name="input_a7dfb8f0181041fcac75b47ffa647167_2_1" id="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_0" class="field-input input-radio" value="choice_0"/><label id="a7dfb8f0181041fcac75b47ffa647167_2_1-choice_0-label" for="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_a7dfb8f0181041fcac75b47ffa647167_2_1">
<span style="font-family: 'book antiqua', palatino;">\(0.p\)</span>
</label>
</div>
<div class="field">
<input type="radio" name="input_a7dfb8f0181041fcac75b47ffa647167_2_1" id="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="a7dfb8f0181041fcac75b47ffa647167_2_1-choice_1-label" for="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_a7dfb8f0181041fcac75b47ffa647167_2_1">
<span style="font-family: 'book antiqua', palatino;">\(\frac{p}{10^n - 1} \) </span>
</label>
</div>
<div class="field">
<input type="radio" name="input_a7dfb8f0181041fcac75b47ffa647167_2_1" id="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="a7dfb8f0181041fcac75b47ffa647167_2_1-choice_2-label" for="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_a7dfb8f0181041fcac75b47ffa647167_2_1">
<span style="font-family: 'book antiqua', palatino;">\(\frac{n}{10^p - 1} \)</span>
</label>
</div>
<div class="field">
<input type="radio" name="input_a7dfb8f0181041fcac75b47ffa647167_2_1" id="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="a7dfb8f0181041fcac75b47ffa647167_2_1-choice_3-label" for="input_a7dfb8f0181041fcac75b47ffa647167_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_a7dfb8f0181041fcac75b47ffa647167_2_1">
<span style="font-family: 'book antiqua', palatino;">\(p - 0.pppp\dots \)</span>
</label>
</div>
<span id="answer_a7dfb8f0181041fcac75b47ffa647167_2_1"/>
</fieldset>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><p>
<span style="font-family: 'book antiqua', palatino;">Is \(0.(p)\) a rational number?</span>
</p>
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Problem 5
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<span style="font-family: 'book antiqua', palatino;">Every real number with a periodic decimal notation is a rational number.</span>
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<span style="font-family: 'book antiqua', palatino;">True or False?</span>
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<span style="font-family: 'book antiqua', palatino;">(If it's true, try proving it. If it's false, find a counterexample.)</span>
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<h2 class="hd hd-2 unit-title">The Proof</h2>
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<p><span style="font-family: 'book antiqua', palatino;">Now for the main event: we will prove \(|\mathbb{N}| < |\mathbb{R}|\).</span></p>
<h4 id="our-strategy."><span style="font-family: 'book antiqua', palatino;">Our Strategy:</span></h4>
<p><span style="font-family: 'book antiqua', palatino;">We’ll focus our attention on the subset <span class="math inline">\([0,1)\)</span> of <span class="math inline">\(\mathbb{R}\)</span>, which consists of real numbers greater or equal to 0 but smaller than 1. Proving <span class="math inline">\(|\mathbb{N}| < |[0,1)|\)</span> is enough for present purposes because we know that <span class="math inline">\(|(0,1]| \leq |\mathbb{R}|\)</span>, and it follows from exercise of Section that <span class="math inline">\(|\mathbb{N}| < |[0,1)|\)</span> and <span class="math inline">\(|(0,1]| \leq |\mathbb{R}|\)</span> together entail <span class="math inline">\(|\mathbb{N}| < |\mathbb{R}|\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">In order to prove <span class="math inline">\(|\mathbb{N}| < |[0,1)|\)</span>, we’ll need to verify each of the following two claims:</span></p>
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<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(|\mathbb{N}| \leq |[0,1)|\)</span></span></p>
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<p><span class="math inline" style="font-family: 'book antiqua', palatino;">\(|\mathbb{N}| \neq |[0,1)|\)</span></p>
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<p><span style="font-family: 'book antiqua', palatino;">The first of these claims is totally straightforward. In fact, we've already looked at a proof of it.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">So all we need to prove is <span class="math inline">\(|\mathbb{N}| \neq |[0,1)|\)</span>. We’ll do so by using a technique that is sometimes called <em>reductio ad absurdum</em> (from the Latin for “reduction to absurdity”). The basic idea is very simple. Suppose you want to prove not-<span class="math inline">\(P\)</span>. You can proceed by assuming its negation (which is not-not-<span class="math inline">\(P\)</span>, or, equivalently, <span class="math inline">\(P\)</span>), and trying to prove a contradiction. If you succeed, you’ve learned that <span class="math inline">\(P\)</span> is false (since no truth entails a contradiction), and therefore that not-<span class="math inline">\(P\)</span> is true, which is what you wanted to show.</span></p>
<h4 id="the-proof-itself" class="unnumbered">The proof itself:</h4>
<p><span style="font-family: 'book antiqua', palatino;">The claim we want to prove here is <span class="math inline">\(|\mathbb{N}| \neq |[0,1)|\)</span>. So we’ll assume <span class="math inline">\(|\mathbb{N}| = |[0,1)|\)</span>, and use it to prove a contradiction. This will show that <span class="math inline">\(|\mathbb{N}| = |[0,1)|\)</span> is false, and therefore that <span class="math inline">\(|\mathbb{N}| \neq |[0,1)|\)</span> is true.</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(|\mathbb{N}| = |[0,1)|\)</span> is the claim that there is a bijection <span class="math inline">\(f\)</span> from <span class="math inline">\(\mathbb{N}\)</span> to <span class="math inline">\([0,1)\)</span>. So let us assume that such an <span class="math inline">\(f\)</span> exists. The first thing to note is that <span class="math inline">\(f\)</span> can be used to make a complete <em>list</em> of real numbers between 0 and 1. The zeroth member of our list is <span class="math inline">\(f(0)\)</span>, the first member of our list is <span class="math inline">\(f(1)\)</span>, and so forth, for each natural number. Since <span class="math inline">\(f\)</span> is a bijection, our list must be complete: every member of <span class="math inline">\([0,1)\)</span> must occur on the list.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">Each real number in <span class="math inline">\([0,1)\)</span> can be represented by a numeral of the form “<span class="math inline">\(0.\delta_0\delta_1\delta_2\ldots\)</span>” where each <span class="math inline">\(\delta_i\)</span> is a digit between “0” and “9”. And since we are excluding duplicate names, each number in <span class="math inline">\([0,1)\)</span> corresponds to a unique such numeral. The listing of <span class="math inline">\([0,1)\)</span> that is induced by <span class="math inline">\(f\)</span> can be represented in the following way:</span></p>
<center>
<figure><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/8683950809f0bf4a1b7f33e4f299a0d7/asset-v1:MITx+24.118x+2T2020+type@asset+block/TheRealNumbers1.png" type="saveimage" target="[object Object]" alt="" /></span><figcaption><span style="font-family: 'book antiqua', palatino;">We are assuming the existence of a function <span class="math inline">\(f\)</span>, which induces a complete list of real numbers in <span class="math inline">\([0,1)\)</span>.</span></figcaption></figure>
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<p><span style="font-family: 'book antiqua', palatino;">I have highlighted certain digits using boldface: the zeorth digit of <span class="math inline">\(f(0)\)</span>, the first digit of <span class="math inline">\(f(1)\)</span>, and, in general, the <span class="math inline">\(n\)</span>th digit of <span class="math inline">\(f(n)\)</span>. These digits form the following sequence: <span class="math display">\[\text{diagonal sequence: } a_0,b_1,c_2,d_3,e_4\ldots\]</span> We will now transform this sequence of digits into an “evil twin”. Let the <strong>evil sequence</strong> be the result of applying the following transformation <span class="math inline">\(\eta\)</span> to each digit <span class="math inline">\(d\)</span> in the diagonal sequence:</span></p>
<p><span style="font-family: 'book antiqua', palatino;"><span class="math display">\[\eta(d) = \begin{cases} \text{``0'', if $d \neq$ ``0''}\\ \text{``1'', if $d =$ ``0''} \end{cases}\]</span> And let the <strong>evil number</strong> be the number whose decimal expansion is “0.” followed by the digits in the evil diagonal sequence. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Suppose, for example, that function <span class="math inline">\(f\)</span> is as as follows:</span></p>
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<figure><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/d133448db999e9228a835b6c09f8a717/asset-v1:MITx+24.118x+2T2020+type@asset+block/TheRealNumbers2.png" type="saveimage" target="[object Object]" alt="" /></span><figcaption><span style="font-family: 'book antiqua', palatino;">An example of the list of real numbers in <span class="math inline">\([0,1)\)</span> that might be induced by <span class="math inline">\(f\)</span>.</span></figcaption></figure>
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<p></p>
<p><span style="font-family: 'book antiqua', palatino;">Then we have the following: <span class="math display">\[\begin{array}{rrl} \text{diagonal sequence:} & \ & 3,\,0,\,1,\,0,\,4,\dots \\ \text{evil sequence:} &\ & 0,\,1,\,0,\,1,\,0,\dots \\ \text{evil number:} &0 \ \ . &0 \ \,1\ \ 0\ \ 1\ \ 0\ \dots \end{array}\]</span> Regardless of what the diagonal sequence turns out to be, the evil number will always be greater than or equal to 0, and smaller than 1. So it will always be a number in <span class="math inline">\([0,1)\)</span>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The climax of our proof is the observation that even though the evil number is in <span class="math inline">\([0,1)\)</span>, it cannot appear anywhere on our list. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">To see this, note that the evil number’s zeroth digit differs from the zeroth digit of <span class="math inline">\(f(0)\)</span>’s decimal expansion, the evil number’s first digit differs from the first digit of <span class="math inline">\(f(1)\)</span>’s decimal expansion, and so forth. (In general, the evil number’s <span class="math inline">\(k\)</span>th digit differs from the <span class="math inline">\(k\)</span>th digit of <span class="math inline">\(f(k)\)</span>’s decimal expansion.) So: <em>the evil number is distinct from every number on our list</em>.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">We have reached our contradiction. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">We began by assuming that <span class="math inline">\(|\mathbb{N}| = |[0,1)|\)</span>. It follows from this assumption that we can generate a complete list of real numbers in <span class="math inline">\([0,1)\)</span>. But we have seen that the evil number is an element of <span class="math inline">\([0,1)\)</span> that cannot be on that list, contradicting our earlier claim that the list is complete. Since <span class="math inline">\(|\mathbb{N}| = |[0,1)|\)</span> entails a contradiction, it is false. So <span class="math inline">\(|\mathbb{N}| \neq |[0,1)|\)</span> is true, which is what we wanted to prove. The infinite size of the real numbers is <em>bigger</em> than the infinite size of the natural numbers. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Amazing!</span></p>
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<h2 class="hd hd-2 unit-title">Some Additional Results</h2>
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<p><span style="font-family: 'book antiqua', palatino;">We have shown that <span class="math inline">\(|\mathbb{N}| < |[0,1)|\)</span>, and therefore that <span class="math inline">\(|\mathbb{N}| < |\mathbb{R}|\)</span>. Notice, however, that this does not immediately settle the question of whether <span class="math inline">\([0,1)\)</span> and <span class="math inline">\(\mathbb{R}\)</span> have the same cardinality, since nothing we’ve proved so far rules out <span class="math inline">\(|\mathbb{N}| < |[0,1)| < |\mathbb{R}|\)</span>. As you’ll be asked to prove below, they do have the same cardinality: <span class="math display">\[|[0,1)| = |\mathbb{R}|\]</span> This is a non-trivial result. It means that just like there are different infinite sets that have the same size as the natural numbers, so there are different infinite sets that have the same size as the real numbers.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The exercises below will ask you to show that the following sets have the same cardinality as the real numbers:<br /></span></p>
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<figure><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/c6a0f4f423b8e55acd90257060a09bfe/asset-v1:MITx+24.118x+2T2020+type@asset+block/TableOfNotation.png" type="saveimage" target="[object Object]" alt="" /></span><figcaption>Some examples of sets with the same cardinality as <span class="math inline">\(\mathbb{R}\)</span>.</figcaption></figure>
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<p><span style="font-family: 'book antiqua', palatino;">As you work through the exercises, you’ll find it useful to avail yourself of the following result: <em>adding natural-number-many members to an infinite set doesn’t change the set’s cardinality,</em>. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">More precisely:</span></p>
<p style="padding-left: 30px;"><span style="font-family: 'book antiqua', palatino;"><strong>No Countable Difference Principle</strong></span><br /><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(|S| = |S \cup A|\)</span>, whenever <span class="math inline">\(S\)</span> is infinite and <span class="math inline">\(A\)</span> is countable.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">(For a set <span class="math inline">\(S\)</span> to be <strong>infinite</strong> is for it to be the case that <span class="math inline">\(|\mathbb{N}| \leq |S|\)</span>; for a set <span class="math inline">\(A\)</span> to be <strong>countable</strong> is for it to be the case that <span class="math inline">\(|A| \leq |\mathbb{N}|\)</span>.)</span></p>
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<strong>No Countable Difference Principle</strong>
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<span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(|S| = |S \cup A|\)</span>, whenever <span class="math inline">\(S\)</span> is infinite and <span class="math inline">\(A\)</span> is countable.</span>
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<span style="font-family: 'book antiqua', palatino;">Recall that the open interval <span class="math inline">\((0,1)\)</span> is the set of of the real numbers <span class="math inline">\(x\)</span> such that <span class="math inline">\(0 &lt; x &lt; 1\)</span>.</span>
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<span style="font-family: 'book antiqua', palatino;">Recall that the closed interval <span class="math inline">\([0,1]\)</span> is the set of the real numbers <span class="math inline">\(x\)</span> such that <span class="math inline">\(0 \leq x \leq 1\)</span>, and that <span class="math inline">\([0,1)\)</span> is the set of the real numbers <span class="math inline">\(x\)</span> such that <span class="math inline">\(0 \leq x &lt; 1\)</span>.</span>
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Problem 4
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<span style="font-family: 'book antiqua', palatino;">For any <span class="math inline">\(a\in\mathbb{R}\)</span> such that <span class="math inline">\(a&gt;0\)</span>, <span class="math inline">\(|[0,a]| = |\mathbb{R}|\)</span>.</span>
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Problem 5
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<span style="font-family: 'book antiqua', palatino;">Just like one can think of the interval <span class="math inline">\([0,1]\)</span> as representing a line of length 1, so one can think of the points in the cartesian product <span class="math inline">\([0,1]\times[0,1]\)</span> as representing a square of side-length 1. (In general, the <strong>Cartesian product</strong> <span class="math inline">\(A\times B\)</span> is the set of pairs <span class="math inline">\(\langle a,b\rangle\)</span> such that <span class="math inline">\(a \in A\)</span> and <span class="math inline">\(b\in B\)</span>.)</span>
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<span style="font-family: 'book antiqua', palatino;">Is it true that there are just as many points on the unit line as there are points on the unit square?: <span class="math display">\[|[0,1]|=|[0,1]\times[0,1]|\]</span></span>
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<h3 class="hd hd-2">Video: Proving that |[0,1]| = |[0,1] x [0,1]|</h3>
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<h3 class="hd hd-3 problem-header" id="0f4e6021c7be40b88a75d0d2144f9437-problem-title" aria-describedby="block-v1:MITx+24.118x+2T2020+type@problem+block@0f4e6021c7be40b88a75d0d2144f9437-problem-progress" tabindex="-1">
Problem 6
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<span style="font-family: 'book antiqua', palatino;">Just like one can think of the points in the cartesian product <span class="math inline">\([0,1]\times[0,1]\)</span> as representing a square of side length 1, so one can think of the points in <span class="math inline">\(\underbrace{[0,1] \times \ldots \times [0,1]}_{\text{$n$ times}}\)</span> as an <span class="math inline">\(n\)</span>-dimensional <em>hypercube</em>.</span>
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<p>
<span style="font-family: 'book antiqua', palatino;"> Is it true that <span class="math inline">\(|[0,1]| = |\underbrace{[0,1] \times \ldots \times [0,1]}_{\text{$n$ times}}|\)</span>?</span>
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<span style="font-family: 'book antiqua', palatino;">(If it's true, try to prove it. If it's false, try to explain why.)</span>
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