Suppose $E$ is a Hausdorff topological space, $A$ is a subset of $E$. Card(A) is the cardinal of $A$. $\bar{A}$ is the closure of $A$. Do we have $Card(\bar{A})\le 2^{2^{Card(A)}}$.
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$\begingroup$ Of course you need Hausdorff, otherwise there are trivial examples where the closure of a singleton is arbitrary large. Even T$_1$ is not enough: for an arbitrary set where open subsets are empty or cofinite, the closure of any infinite subset is everything. $\endgroup$– YCorCommented Oct 29, 2020 at 9:32
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$\begingroup$ @YCor. Thanks. If E is also Hausdorff, is the conclusion valid? $\endgroup$– CheskiCommented Oct 29, 2020 at 9:38
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I'll assume $E$ Hausdorff otherwise there are trivial counterexamples. Then this fact holds.
We can suppose $A$ is dense. Then the map $\Phi:E\to 2^{2^A}$ mapping $x$ to the set of $P\subset A$ such that $x\in\overline{P}$ is injective: this entails the result.
Indeed, suppose $x_1\neq x_2$. So there are disjoint $U_1,U_2$ open in $E$ such that $x_i\in U_i$. Write $X_i=U_i\cap A$. Then $x_i\notin\overline{X_j}$ for $\{i,j\}=\{1,2\}$. If by contradiction $x_i\notin \overline{X_i}$, there exists $V_i$ open neighborhood of $x_i$ in $E$ such that $V_i\cap X_i=\emptyset$. That is, $V_i\cap U_i\cap A$ is empty. By density of $A$, $V_i\cap U_i$ is empty. But $x_i\in V_i\cap U_i$, contradiction.
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2$\begingroup$ And (assuming $A$ is infinite) the bound is attained by the Cantor space $2^{2^A}$ which has a dense subset with the same cardinality as $A$, e,g., the set of all functions $f:2^A\to2$ such that $f(x)$ depends on only finitely many coordinates of $x\in2^A$. $\endgroup$– bofCommented Oct 29, 2020 at 11:39