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\title{\textbf{A problem}}
\maketitle

[Problem] Let $A=\{$functions from $(0,1)$ to $\mathbb{R}\}$, and $B=\{$functions from $(0,1)$ continuously to $\mathbb{R}\}$. Show that (a) $A\not\approx\mathbb{R}$, and (b) $B\approx\mathbb{R}$.

\begin{proof}
We now show(a). Assume that $A\approx (0,1)$. Let $F:(0,1)\to A$ be such a bijection. Define a function $g:(0,1)\to \mathbb{R}$ such that for $c\in (0,1)$, 
\begin{align}
 g(c)=
  \begin{cases}
    0,  &\text{if $(F(c))(c)=1 $; } \\
    1,    &\text{otherwise.}
  \end{cases}
\end{align}

(This is the "diagonal argument" on continuum.) We know that $g\in A=F((0,1))$. For $x\in (0,1)$, since $g(x)\ne (F(x))(x)$, we have that $g\ne F(x)$. This means $g\notin F((0,1))$, a contradiction.

Therefore, $A\not\approx\mathbb{R}$. 


Secondly, we prove (b). To describe a function from $(0,1)$ continuously to $\mathbb{R}$, we only need to know its values on rationals. Moreover, the set 
\begin{align}
C(\mathbb{Q}\cap (0,1),\mathbb{R})\subset \mathbb{R}^{\mathbb{Q}\cap (0,1)}.\notag
\end{align}
Note that restriction sets up a bijection between $B$ and $C(\mathbb{Q}\cap(0,1),\mathbb{R})$. So we have 
\begin{align}
|B|\leq |\mathbb{R}^{\mathbb{Q}\cap (0,1)}| = \mathfrak{c}.\notag
\end{align}
We also know that $|B|\geq\mathfrak{c}$ because 
\begin{align}
B\supset \bigcup_{r\in\mathbb{R}} \{f:f(x)\equiv r\}.\notag
\end{align}
Therefore, $|B|=\mathfrak{c}$, i.e. $B\approx\mathbb{R}$.
\end{proof}


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