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There is a statement as follows:

If a Hausdorff (regular, Tychonoff) space $X$ has countable pseudocharacter and countable tightness, then the closure of any set $Y\subset X$ of cardinality $\le \mathfrak c$, has cardinality $\le \mathfrak c$.

My work: I could prove that $|\overline{Y}| \le 2^\mathfrak c$ only by the following inequality $|X| \le 2^{d(X)\psi(X)} \le 2^\mathfrak c$ for any regular space $X$.

Is the above statement right? If it is, how could I prove it?

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  • $\begingroup$ @HennoBrandsma yes, i have done it. $\endgroup$
    – Paul
    Apr 6 '17 at 9:51
  • $\begingroup$ There is many results on this topic (I mean evaluating the cardinality of a topological space via other cardinal invariants), see the survey paper of Hodel: (citeseerx.ist.psu.edu/viewdoc/…). In this paper of Hodel there is an Example on page 3. Maybe it will be helpful. $\endgroup$ Apr 6 '17 at 10:24
  • $\begingroup$ @TarasBanakh the Katetov extension was exactly my example to show the necessity of regularity as opposed to Hausdorff. $\endgroup$ Apr 6 '17 at 14:14
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Copying my answer to the almost duplicate question on math stackexchange:

Assume $X$ is $T_3$ and has countable tightness and pseudocharacter. We will show that $|X|\le d(X)^\omega$, which will imply what you want using $d(\overline{Y}) \le |Y| = 2^\omega$.

Let $D$ be a dense subset of size $d(X)$. As each $x \in X$ is in $\overline{D}$, and $X$ has countable tightness we can choose for each $x \in X$ a countable subset $D_x$ of $D$ such that $x \in \overline{D_x}$.

As $X$ has countable pseudocharacter, and is regular, we can pick for each $x$ a countable sequence $(U(x)_n)_n$ of open sets such that $$\{x\} = \bigcap_n \overline{U(x)_n}\text{.}$$

Now define $$f: X \rightarrow ([D]^\omega)^\omega \text{ by } f(x) = (U_n(x) \cap D_x)_{n \in \omega}$$

Claim: $f$ is 1-1. For suppose $x \neq y$. Then there exists an $n$ such that $y \notin \overline{U_n(x)}$. As clearly for any $n$, $y \in \overline{U_n(y) \cap D_y}$ ,we see that for that first $n$: $$y \notin \overline{U_n(x) \cap D_x} \subseteq \overline{U(x)_n} \text{ but } y \in \overline{U_n(y) \cap D_y}$$

meaning that $f(x)_n \neq f(y)_n$ (the closures of these sets are different so the sets are different too) so $f(x) \neq f(y)$.

As $|([D]^\omega)^\omega| = |D^\omega| = d(X)^\omega$, we have that $|X| \le d(X)^\omega$ for those spaces.

Analysing the proof we see that in fact that for regular spaces $X$:

$$|X| \le d(X)^{t(X)\psi(X)}$$

using sequences of length $\psi(X)$ of subsets of $D$ of size $t(X)$.

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Since $X$ is countably tight, closure of $Y$ is the union of closures of its countable subsets. So it is enough to show that the closure of a countable set $A$ has size $\leq \mathfrak{c}$. To each point $x$ in the closure of $A$ associate a sequence $S_x$ of countable subsets of $A$ as follows. Choose a sequence of open sets $\langle U_{n, x} : n \geq 1\rangle$ whose intersection is $\{x\}$ (using countable pseudocharacter) and let $S_x = \langle U_{n, x} \cap A : n \geq 1 \rangle$. It is easy to check that different points in the closure of $A$ are associated with different sequences. Since the cardinality of the set of countable sequences of subsets of a countable set is $\leq \mathfrak{x}$, it follows that the closure of $A$ has size $\leq \mathfrak{c}$.

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  • $\begingroup$ Easy to check how? I used $T_3$ in my proof. I'd like to see your argument, as my assumption could have been redundant. $\endgroup$ Apr 5 '17 at 7:33
  • $\begingroup$ Update: I think regularity is needed, I added a Hausdorff example to my answer linked to in my answer on math.stackexchange. Use $Y = \omega$ in that example and we have $\overline{Y} = 2^\mathfrak{c}$ $\endgroup$ Apr 5 '17 at 7:49

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