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Math Help - Point-Set Topology

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    MHF Contributor Drexel28's Avatar
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    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?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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:
    The answer is no.

    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.
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