1. ## Point-Set Topology

Bruno's posts got me in the mood (hint hint) for this.

Problem:

Let $\mathbb{R}^{\mathbb{R}}=\left\{f\mid f:\mathbb{R}\to\mathbb{R}\right\}$. Is it possible to define a topology on $\mathcal{P}\left(\mathbb{R}^{\mathbb{R}}\right)$ which turns it into a compact metric space?

2. Originally Posted by Drexel28
Bruno's posts got me in the mood (hint hint) for this.

Problem:
The answer is no. This might be kind of an esoteric question, so if anyone doesn't post by tomorrow I'll just post up my solution.

3. Originally Posted by Drexel28
The answer is no. This might be kind of an esoteric question, so if anyone doesn't post by tomorrow I'll just post up my solution.
As promised:

Spoiler:

Lemma: Every quasi-totally bounded (instead of finite, there is a countable net) metric space $M$ is separable.

Proof: For every $n\in\mathbb{N}$ let $\mathcal{A}_n$ be the guaranteed countable $\tfrac{1}{n}$-net and let $\mathcal{A}=\bigcup_{n=1}^{\infty}\mathcal{A}_n$. Clearly $\mathcal{A}$ being the countable union of countable sets is countable. Also, we claim that it is dense in $M$. To see this let $m\in M$ and $\varepsilon>0$ be given. By the Archimedean principle there exist some $n\in\mathbb{N}$ such that $\frac{1}{n}<\varepsilon$. So, then since $\left\{B_{\frac{1}{n}}(a)\right\}_{a\in\mathcal{A} _n}$ cover $M$ it must be that $x\in B_{\frac{1}{n}}(a)$ for some $a\in\mathcal{A}_n\subseteq\mathcal{A}$. Thus, for that $a$ we have that $d(x,a)<\frac{1}{n}<\varepsilon$ from where the conclusion follows. $\blacksquare$

It follows since every compact metric space is totally bounded that every compact metric space is separable.

Also:

Lemma: Let $X$ be a $T_1$ space with $\mathfrak{D}\subseteq X$ dense. Then, $\text{card }X\leqslant 2^{2^{\text{card }\mathfrak{D}}}=2\text{ card }\mathcal{P}\left(\mathcal{P}\left(\mathfrak{D}\ri ght)\right)$

Proof: For a fixed $x\in X$ define $\mathfrak{N}_x$ to be the set of all neighborhoods of $x$ and define $\Upsilon:X\to\mathcal{P}\left(\mathcal{P}\left(\ma thfrak{D}\right)\right):x\mapsto\left\{N\cap\mathf rak{D}:N\in\mathfrak{N}_x\right\}$. To see this is an injection we merely note that if $x\ne y$ then $X$ being $T_1$ guarantees there are neighboorhoods $U\in\mathfrak{N}_x$ and $V\in\mathfrak{N}_y$ such that $y\notin U$ and $x\notin V$. In particular since every neighborhood of $y$ contains $y$ we have that $U\notin\mathfrak{N}_x$ and so $U\cap N\notin\Upsilon(y)$ and so $\Upsilon(x)\ne\Upsilon(y)$. It follows that $\Upsilon$ is an injection from where the conclusion follows. $\blacksquare$

Thus, either looking here, or noticing that $\text{card }\mathbb{R}^{\mathbb{R}}=\mathfrak{c}^{\mathfrak{c }=\left(2^{\aleph_0}\right)^{\mathfrak{c}}=2^{\ale ph_0}\cdot\mathfrak{c}}=2^{\mathfrak{c}}=\text{car d }\mathcal{P}(\mathbb{R})=\text{card }\mathcal{P}\left(\mathcal{P}\left(\mathbb{N}\righ t)\right)$ and so $\text{card }\mathcal{P}\left(\mathbb{R}^{\mathbb{R}}\right)=\ text{card }\mathcal{P}\left(\mathcal{P}\left(\mathcal{P}\lef t(\mathbb{N}\right)\right)\right)>\text{card }\mathcal{P}\left(\mathcal{P}\left(\mathbb{N}\righ t)\right)$ and so by the above lemma $\mathcal{P}\left(\mathbb{R}^{\mathbb{R}}\right)$ cannot be turned into a separable $T_1$ space and thus not into a compact metric space.