On the cardinality of a $T_1$ space

Question

I have a question about the "denomination" of a space $(X,\mathcal{T})$ that satisfies the first axiom of separation. I found that the $T_1$ spaces are also called accessible and I wonder if that name appeal to the cardinality of the set. I have read that the fact that $(X,\mathcal{T})$ satisfies certain separation axioms restricts the cardinality of the space;

---

for example, if $(X,\mathcal{T})$ is a $T_0$ space and a second-countable space then $card(X)\le \aleph_0$. Where $card(A)$ means "the cardinality of the set $A$".

---

Then my question is, why we can call a $T_1$ space as accessible? Is this related to the inaccessible cardinals?

I have search on books of topology and I have not found an answer

---

Thanks in advance. I'm not sure which tag put on this question. I appreciate if someone corrects me.

---

Ok, the reals with the usual topology is a good counterexample about I write before. However the question still stands!!

Answer 1

It's certainly not true that a second countable $T_0$ space $X$ has $|X| \le \aleph_0$ (I'll use the more standard $|X|$ for cardinality). The reals are already a counterexample.

But the map $f: X \rightarrow \mathscr{P}(\mathcal{B})$, where $\mathcal{B}$ is a fixed base for $X$ ,and $\mathscr{P}(A)$ denotes the power set of $A$, defined by

$$f(x) = \{B \in \mathcal{B}: x \in B\}$$

is 1-1: suppose $X \neq y$ then by $T_0$-ness there is some open set $O$ such that (say) $x \in O, y \notin O$. Then as $\mathcal{B}$ is a base there is a $B \in \mathcal{B}$ such that $X \in B \subseteq O$. But then $y \notin B$ as well (as $y \notin O$), and $x \in B$ which shows $B \in f(x)$, $B \notin f(y)$, so the subsets $f(x), f(y)$ of $\mathcal{B}$ are different. So $f$ is 1-1. so by 1-1 ness of $f$ we can bound the cardinality of $X$:

$$|X| \le |\mathscr{P}(\mathcal{B})| = 2^{|\mathcal{B}|}$$

Now let $w(X)$, the weight of $X$ denote the minimal size for a base for $X$ (So $X$ second countable iff $w(X) \le \aleph_0$).

Hence, a $T_0$ space that is second countable, has size at most $2^{\aleph_0} =\mathfrak{c} = |\mathbb{R}|$. So $\mathbb{R}$ already realises this maximum.

Some (mild) separation axioms is needed because $X$ is the indiscrete topology can have any size we like, so there is no upperbound on its size.