Consider a space $X$ with two topologies $\tau_1$ and $\tau_2$. It is easy to see that nets remain convergent if the topology is made coarser, i.e.
$$\tau_1 \subseteq \tau_2 \quad \Rightarrow \quad \Big\{ x_\alpha \overset{\tau_2}{\longrightarrow} x \, \Rightarrow \, x_\alpha \overset{\tau_1}{\longrightarrow} x \Big\}.$$
My question is: does the converse hold? i.e. is it true that
$$\tau_1 \subseteq \tau_2 \quad \iff \quad \Big\{ x_\alpha \overset{\tau_2}{\longrightarrow} x \, \Rightarrow \, x_\alpha \overset{\tau_1}{\longrightarrow} x \Big\}$$
? I think this is related to whether a topology is fully determined by the convergence of nets, but I couldn't find a clear answer.
Best Answer
I would like to attempt an answer motivated by a comment by GEdgar.
I want to show that
$$ \Big\{ x_\alpha \overset{\tau_2}{\longrightarrow} x \, \Rightarrow \, x_\alpha \overset{\tau_1}{\longrightarrow} x \Big\} \quad \Rightarrow \quad \tau_1 \subseteq \tau_2.$$
So consider $U \in \tau_1$. According to the theorem in this post, this is equivalent to saying that
$$\forall x \in U \, \text{ and } \, \forall x_\alpha \overset{\tau_1}{\longrightarrow} x, \exists \beta \, \text{ such that } \, x_\alpha \in U, \, \, \forall \alpha \geq \beta \, . $$
By assumption, the set of nets that $\tau_2$-converge to $x$ is included in the set of nets that $\tau_1$-converge to $x$. Thus the above statement also holds with $\tau_2$:
$$\forall x \in U \, \text{ and } \, \forall x_\alpha \overset{\tau_2}{\longrightarrow} x, \exists \beta \, \text{ such that } \, x_\alpha \in U, \, \, \forall \alpha \geq \beta \, . $$
In conclusion, $U \in \tau_2$.
Is the argument correct?