Motivating the terminology of weaker vs stronger topology

Question

Let $X$ be a set, $\tau_1,\tau_2$ two topologies on $X$, and consider the following statements

1. $\tau_1\subseteq \tau_2$ (i.e $\tau_1$ is coarser/weaker than $\tau_2$, or that $\tau_2$ is finer/stronger than $\tau_1$)
2. For every net $\langle x_i\rangle_{i\in I}$ in $X$ and $x\in X$, if $x_i\to x$ relative to the topology $\tau_2$ then $x_i\to x$ relative to the topology $\tau_1$.

It is clear to me that $(1)$ implies $(2)$. My question is whether the converse is also true, because then this would seem like a nice justification for the terminology "weak/strong" topology in the context of topological vector spaces.

Answer 1

The converse is true: let´s prove $\neg1\implies\neg2$. If you have a set $U$ in $\tau_1$ but not in $\tau_2$, you can pick a point $x\in U$ which isn´t in the interior of $U$ respect to the topology of $\tau_2$. Thus, if $(B_i)_{i\in I}$ is a base of nhoods of $x$ in $\tau_2$ ordered by reverse inclusion, for each $i$ you can pick a point $x_i\in B_i\setminus U$. So the net $(x_i)_{i\in I}$ converges to $x$ in $\tau_2$ but it doesn´t in $\tau_1$, because no point of the net is in $U$.

Answer 2

Suppose (2) holds. Let $O \in \tau_1$. Then $X\setminus O$ is closed in $\tau_1$, and it's also closed in $\tau_2$: let $x$ be in the closure (w.r.t $\tau_2$) of $X\setminus O$, so there is a net $(x_i)_i$ of points from $X\setminus O$ such that $x_i \to x$ in $\tau_2$. By $(2)$ we also have that $x_i \to x$ in $\tau_1$, so that $x$ lies in $X\setminus O$ as that set is already closed in $\tau_1$. This shows that $X\setminus O$ is also closed under $\tau_2$ so $O \in \tau_2$ and we're done.