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?



A result on spaces with countable pseudocharacter and countable tightness
Copying [my answer to the almost duplicate question on math stackexchange](https://math.stackexchange.com/a/2217145/4280):


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 $t(X)$. 


