$\kappa\psi (x,X)\leq \psi (x,X)$ The $\kappa$-pseudocharacter $\kappa\psi (x,X)$ of a space $X$ at a point
$x\in X$ is the smallest infinite cardinal number $\tau$ such that
there exists a family $\gamma$
 of $\kappa$-sets in $X$ such that
$\{x\} = \bigcap\gamma$.
For every topological group we have following inequality:

$$\kappa\psi (x,X)\leq \psi (x,X)$$

But how about topological spaces, Is it true for every topological spaces?

Closure of any open set is called $\kappa$-set.
 A: $\newcommand{\cl}{\operatorname{cl}}$Here’s a Hausdorff counterexample. Let $$X=\Big(\omega_1\times(\omega+1)\Big)\cup\{p\}\;,$$ where $p$ is a point not in $\omega_1\times(\omega+1)$. Basic open nbhds of $p$ are the sets
$$B(\alpha,n)=\{p\}\cup\{\langle\xi,k\rangle\in\omega_1\times\omega:\xi>\alpha\text{ and }k\ge n\}$$
for $\langle\alpha,n\rangle\in\omega_1\times\omega$. Clearly $\{p\}=\bigcap_{n\in\omega}B(0,n)$, so $\psi(x,X)=\omega$. Suppose now that $U$ is an open nbhd of $p$. Then $p\in B(\alpha,n)\subseteq U$ for some $\langle\alpha,n\rangle\in\omega_1\times\omega$, and clearly $$(\alpha,\omega_1)\times\{\omega\}\subseteq\cl B(\alpha,n)\subseteq\cl U\;.$$ 
Let $\{\cl U_k:k\in\omega\}$ be a countable family of $\kappa$-sets containing $p$. For each $k\in\omega$ there is a pair $\langle\alpha_k,n_k\rangle\in\omega_1\times\omega$ such that $B(\alpha_k,n_k)\subseteq U_k$. Let $\alpha=\sup\{\alpha_k:k\in\omega\}$; then 
$$(\alpha,\omega_1)\times\{\omega\}\subseteq\bigcap_{k\in\omega}\cl B(\alpha_k,n_k)\subseteq\bigcap_{k\in\omega}\cl U_k\;,$$
and it follows that $\kappa\psi(x,X)>\omega$.
