Separation properties of a topological space vs. characteristics of the continuum Suppose that a set $X$ has a topology $\mathcal{T}$. Then $$\mathcal{T}\ \text{is T}_1\Rightarrow|\mathcal{T}|\geq|X|.$$ I'm curious about implications in the opposite direction, possibly assuming the negation of the continuum hypothesis.
I would also be interested in implications in the same direction which are finer than allowed with CH, i.e. if $|(|X|,|\mathcal{P}(X)|)|>0$ can hold for infinite sets $X$.
 A: The converse absolutely doesn't hold. Just take an infinite $T_1$ topological space $(X, \mathcal{T})$, adjoin a point $y \notin X$ to form $Y = X \cup \{y\}$, and define
$$\mathcal{T}' = \{\emptyset\} \cup \{\mathcal{U} \cup \{y\} : \mathcal{U} \in \mathcal{T}\}.$$
Then $\mathcal{T}'$ is a topology on $Y$ that is not $T_1$ (the point $y$ cannot be separated from any other point with an open set). But, $|X| = |Y|$, since $X$ is infinite, and $|\mathcal{T}'| = |\mathcal{T}|$, since we can put $\mathcal{T}' \setminus \{\emptyset\}$ in bijective correspondence with $\mathcal{T}$, and by the quoted result, $\mathcal{T}$ is infinite.
A: Let $\langle X,\tau\rangle$ be any infinite space, and let $I=\{0,1\}$ with the indiscrete topology. Then $X\times I$ has the same cardinality as $X$, and the product topology on $X\times I$ has the same cardinality as $\tau$, since the open sets in the product are the sets of the form $U\times I$ for $u\in\tau$, but the product is not even $T_0$. Thus, any combination of cardinalities of $X$ and $\tau$ that is possible at all is possible for a space that is not even $T_0$.
