<p>It's actually more trivial. </p>

<ol>
<li><p>Suppose that $s$ occurs in the enumeration. </p></li>
<li><p>Then it occurs at some specific index. Let's call this index $n$. This means that $s = s_n$. </p></li>
<li><p>But this is impossible, because $s$ and $s_n$ differ in the $n$th digit.</p></li>
</ol>

CC BY-SA 3.0
2015-01-17T04:33:49.287Z
2015-01-17T04:33:49.287Z
Alexey Romanov
<p>I struggled to understand the <a href="http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument" rel="nofollow noreferrer">Cantor's diagonal argument</a>, but I have some problems comprehending the following:</p>
<blockquote>
<p>By construction, <span class="math-container">$s$</span> differs from each <span class="math-container">$s_n$</span>, since their <span class="math-container">$n^{th}$</span> digits differ (highlighted in the example). <strong>Hence, <span class="math-container">$s$</span> cannot occur in the enumeration.</strong></p>
</blockquote>
<p>Can someone explain why the built sequence <span class="math-container">$s$</span> cannot occur in the enumeration? I can imagine it somehow recursively that if we add it to <span class="math-container">$s_n$</span>, we can pick another sequence with the same construction pattern and we can repeat these steps forever. However, I want to know if this is sufficient as a proof or there is some kind of other explanation.</p>

<set-theory>
Explanation of Cantor's diagonal argument?
Ivaylo Toskov
# Explanation of Cantor's diagonal argument?

(Asked by: username_0 on 2015-01-17)

I struggled to understand the [Cantor's diagonal argument](http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument), but I have some problems comprehending the following:

> By construction, $s$ differs from each $s\_n$, since their $n^{th}$ digits differ (highlighted in the example). **Hence, $s$ cannot occur in the enumeration.**

Can someone explain why the built sequence $s$ cannot occur in the enumeration? I can imagine it somehow recursively that if we add it to $s\_n$, we can pick another sequence with the same construction pattern and we can repeat these steps forever. However, I want to know if this is sufficient as a proof or there is some kind of other explanation.

**Tags:** set-theory

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## 1 answers

### Answer by username_1 on 2015-01-17. Score: 1 | Accepted answer

It's actually more trivial.

1. Suppose that $s$ occurs in the enumeration.
2. Then it occurs at some specific index. Let's call this index $n$. This means that $s = s\_n$.
3. But this is impossible, because $s$ and $s\_n$ differ in the $n$th digit.


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