questions about point set topology

Question

Let $X$ be a topological space. $\tau$ be a topology on $X$.

A base $B$ of $\tau$ is a collection of subsets of $X$ such that

1. $\forall b \in B$, $b \in \tau$

2. $X$ is covered by the base elements. (aka their union would be $X$)

3. if $b_1$, $b_2 \in B$, and $\exists p \in b_1\cap b_2$ then $\exists b_3 \in B$ where $p \in b_3 \subseteq b_1\cap b_2$

^straight from class notes and double checked at many places

Then let $A = \left\{ X\right\}$. Obviously

1. $X \in \tau$

2. $X$ covers $X$

3. is true vacuously since there is only one element in $A$, which is $X$ so there will be no intersection with other base elements.

Therefore $A$ is a base.

Then my question is since $A$ is countable, then $X$ is second countable. Then since we can construct something like this for all topological spaces, therefore all topological spaces are second countable?

Intuitively, this does not make sense because every open subsets should be the union of some base elements.

I know I am wrong, but not sure how I screwed up.

Answer 1

I think a point of confusion here is that a collection $B$ of subsets of $X$ satisfying the conditions you listed is a basis for $\textit{some}$ topology on $X$. We cannot start with a topology $\tau$ on $X$ then take such a collection $B$ and hope that it is a basis for $\tau$. As David Wheeler points out in the comments, your set $A$ is a basis solely for the indiscrete topology on a set $X$.

If you're given a topology $(X,\tau)$, and you want to find a basis for $\tau$, then you need a collection of open sets $B$ in $X$ such that for any open set $U \subseteq X$ there exist $B_i \in B$ with $U= \bigcup_i B_i$.

Answer 2

Your choice of $A = \{X\}$ is indeed a base for the indiscrete topology, a.k.a the trivial topology $\tau = \{\emptyset,X\}$.

If a collection satisfies those properties, then it is a base for some topology. That doesn't mean that it is the base for an arbitrary topology.

In order for a base $B$ to the base for a topology $\tau$, it must generate all the sets in $\tau$ in addition to the rules you listed.