Merge pull request #798 from gz101/patch-3

Minor typo in equation

Author: boazbk

lec_02_representation.md

@@ -429,7 +429,7 @@ We start by proving  [sequencestostrings](){.ref} which is really the heart of [
 
 
 
-![We construct a function $\overline{d}$ such that $\overline{d} \neq StF(x)$ for every $x\in \{0,1\}^*$ by ensuring that $\overline{d}(n(x)) \neq StF(x)(n(x))$ for every $x\in \{0,1\}^*$ with lexicographic order $n(x)$. We can think of this as building a table where the columns correspond to numbers $m\in \N$ and the rows correspond to $x\in \{0,1\}^*$ (sorted according to $n(x)$). If the entry in the $x$-th row and the $m$-th column corresponds to $g(m))$ where $g=StF(x)$ then $\overline{d}$ is obtained by going over the "diagonal" elements in this table (the entries corresponding to the $x$-th row and $n(x)$-th column) and ensuring that $\overline{d}(x)(n(x)) \neq StF(x)(x)$. ](../figure/diagonalization.png){#diagrealsfig   }
+![We construct a function $\overline{d}$ such that $\overline{d} \neq StF(x)$ for every $x\in \{0,1\}^*$ by ensuring that $\overline{d}(n(x)) \neq StF(x)(n(x))$ for every $x\in \{0,1\}^*$ with lexicographic order $n(x)$. We can think of this as building a table where the columns correspond to numbers $m\in \N$ and the rows correspond to $x\in \{0,1\}^*$ (sorted according to $n(x)$). If the entry in the $x$-th row and the $m$-th column corresponds to $g(m))$ where $g=StF(x)$ then $\overline{d}$ is obtained by going over the "diagonal" elements in this table (the entries corresponding to the $x$-th row and $n(x)$-th column) and ensuring that $\overline{d}(n(x)) \neq StF(x)(n(x))$. ](../figure/diagonalization.png){#diagrealsfig   }
 
 
 __Warm-up: "Baby Cantor".__ The proof of [sequencestostrings](){.ref} is rather subtle. One way to get intuition for it is to consider the following finite statement "there is no onto function $f:\{0,\ldots,99\} \rightarrow \{0,1\}^{100}$". Of course we know it's true since the set $\{0,1\}^{100}$ is bigger than the set $[100]$, but let's see a direct proof. For every $f:\{0,\ldots,99\} \rightarrow \{0,1\}^{100}$, we can define the string $\overline{d} \in \{0,1\}^{100}$ as follows: $\overline{d} = (1-f(0)_0, 1-f(1)_1 , \ldots, 1-f(99)_{99})$. If $f$ was onto, then there would exist some $n\in [100]$ such that $f(n) =\overline{d}$, but we claim that no such $n$ exists.  Indeed, if there was such $n$, then the $n$-th coordinate of $\overline{d}$ would equal $f(n)_n$ but by definition this coordinate equals $1-f(n)_n$. See also a ["proof by code"](https://trinket.io/python/4cff7e58f4) of this statement.