# Convergence of a sequence in terms of sub-basis of a topological space

Let $$X$$ be a topological space and $$S ⊆ 2^X$$ a sub-basis for the topology of $$X$$. Show that a sequence $$x_1, x_2, . . .$$ in $$X$$ converges to a point $$x ∈ X$$ if and only if for every $$A ∈ S$$ containing $$x$$, there exists an $$n_0 ∈ \mathbb N$$ such that for all $$n ≥ n_0$$, we have $$x_n ∈ A$$.

My attempt:

Definition: Let $$X$$ be a topological space. A sequence $$x_1,x_2,...$$ in $$X$$ converges to $$x$$ in $$X$$ if and only if for every open neighborhood $$U$$ of $$x$$, there exists $$n_0 \in \mathbb N$$, such that for all $$n \geq n_0$$, we have $$x_n \in U$$.

1. $$=>$$: since $$S$$ is a sub-basis for the topology of $$X$$, then $$X = U_{A_i \in S} A_i$$. So if the sequence $$x_1,x_2,...$$ in $$X$$ converges to $$x$$ in $$X$$, then $$x_1,x_2,...$$ in $$U_{A_i \in S} A_i$$, hencr each $$A_i$$ is an open neighborhood of $$x$$, so $$x_n \in A$$.

2. $$<=$$: if $$x_n \in A$$ then $$x_n \in U_{A_i \in S} A_i$$, then $$x_1, x_1,...$$ in $$U_{A_i \in S} A_i$$, so for each open $$A_i$$ of $$x_n$$, there exists $$n_0 \in \mathbb N$$ such that for all $$n \geq n_0$$, we have $$x_1,x_2,...$$ in $$X$$ converges to $$x$$ in $$X$$.

Combing $$1$$ and $$2$$,I'd get the required result. Is my attempt correct?

$$\implies$$

Let $$(x_n)_n$$ converge to $$x$$ and let $$x\in A\in\mathcal S$$.

The topology is generated by $$\mathcal S$$ hence $$A$$ is an open neighborhood of $$x$$.

So some $$n_0$$ exists with $$n\geq n_0\implies x_n\in A$$.

$$\impliedby$$

Let it be that for every $$A\in\mathcal S$$ there is some integer $$m$$ with $$n\geq m\implies x_n\in A$$, and let $$U$$ be an open neighborhood of $$x$$.

Then a finite sequence $$A_1,\dots,A_k$$ exists with $$A_i\in\mathcal S$$ and $$x\in\bigcap_{i=1}^kA_i\subseteq U$$ because $$\mathcal S$$ is a subbase of the topology.

For each $$i\in\{1,\dots,k\}$$ there is some integer $$n_i$$ with $$n\geq n_i\implies x_n\in A_i$$.

Now let $$n_0=\max(\{n_1,\dots,n_k\})$$.

Then $$n\geq n_0\implies x_n\in\bigcap_{i=1}^kA_i\subseteq U$$.