Suppose X and Y are Hausdorff spaces and $f: X \to Y$ is a continous map. Prove that cardinality of $\overline{f(X)}$ not more than $2^{U}$, where $U$ -- family of open subsets of X.
Is it right or need to take $2^{2^{U}}$?
Hausdorff spaces and cardinality of $\overline{f(X)}$
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general-topology
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0It's right. Look for an injection $\overline{f(X)} \hookrightarrow \mathcal{P}(U)$. – 2017-01-14
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0It's clear that exists $f(X) \hookrightarrow \mathcal{P(U)}$, but what can I say about $\partial{f(X)}$? – 2017-01-14
1 Answers
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It's not important whether $X$ is Hausdorff, only the Hausdorff property of $Y$ is needed. For $y\in \overline{f(X)}$, define
$$\mathscr{F}(y) = \{ f^{-1}(V) : y \in V, \: V \text{ open}\}.$$
By continuity of $f$, $\mathscr{F}(y)$ is a family of open subsets of $X$, i.e. $\mathscr{F}(y) \in \mathcal{P}(U)$. Since we assumed $y \in \overline{f(X)}$, we have $V\cap f(X) \neq \varnothing$ and hence $f^{-1}(V) \neq \varnothing$ for every open $V$ with $y\in V$. So $\varnothing \notin \mathscr{F}(y)$ for all $y\in \overline{f(X)}$.
To conclude, we show that $\mathscr{F}$ is injective. Let $y_1, y_2$ be different points in $\overline{f(X)}$. Since $Y$ is Hausdorff, there is an open $V \subset Y$ with $y_1 \in V$ and $y_2 \notin \overline{V}$. Then $f^{-1}(V) \in \mathscr{F}(y_1) \setminus \mathscr{F}(y_2)$: if $W$ is an open set containing $y_2$, then $W\setminus \overline{V}$ is also an open set containing $y_2$, and
$$\varnothing \neq f^{-1}(W\setminus \overline{V}) = f^{-1}(W) \setminus f^{-1}(\overline{V}) \subset f^{-1}(W) \setminus f^{-1}(V),$$
whence $f^{-1}(W) \neq f^{-1}(V)$.
Thus $\mathscr{F} \colon \overline{f(X)} \to \mathcal{P}(U)$ is an injective map, from which $\operatorname{card} \overline{f(X)} \leqslant \operatorname{card} \mathcal{P}(U) = 2^{\operatorname{card} U}$ follows.