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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?

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Your attempt is not 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 $n\geq n_0\implies x_n\in\bigcap_{i=1}^kA_i\subseteq U$.

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