Convergence and Comparison of Topology

Question

Let $(X,\mathcal{T}_1)$ and $(X,\mathcal{T}_2)$ be a topological space endowed with two different topologies. If any convergent net $\{x_v\}$ in $(X,\mathcal{T}_1)$ is convergent in $(X,\mathcal{T}_2)$, does it imply that $\mathcal{T}_1\supseteq\mathcal{T}_2$?

Answer 1

The question can be reworded in this way: suppose the identity mapping maps $\mathcal T_1$-convergent nets to $\mathcal T_2$-convergent nets. Does it follow that it is a continuous map $(X,\mathcal T_1)\to (X,\mathcal T_2)$?

This is obviously true if all the $\mathcal T_1$-limits of a given net are also its $\mathcal T_2$-limits (even if there are more $\mathcal T_2$-limits; it is enough to consider nets of $\mathcal T_1$-neighbourhoods), but in general, this need not be the case.

This is a (superficially) special case of another question asked here. In particular, the answer is yes if $\mathcal T_2$ is Hausdorff (or $T_2$, pardon the pun). In fact, I believe this is true even if $\mathcal T_2$ is just $T_1$.

To see this, take any $\mathcal T_1$-convergent net $(x_i)_{i\in I}$ and let $x$ be its $\mathcal T_1$-limit. We need to show that $x$ is also its $\mathcal T_2$-limit. Consider the net $(y_{i,j})_{i\in I,j\in \{0,1\}}$ where $y_{i,0}=x_i$ and $y_{i,1}=x$. Then clearly $(y_{i,j})_{i,j}$ is still a net $\mathcal T_1$-convergent to $x$, and if $\mathcal T_2$ is $T_1$, then it is not $\mathcal T_2$-convergent to any point other than $x$ (because it contains a cofinal net constant at $x$). Thus, since it is $\mathcal T_2$-convergent, it is $\mathcal T_2$-convergent to $x$. But then it follows that its cofinal subnet $(x_i)_i$ is also convergent to $x$, and we are done.

The answer is no in general: directly lifting the example given there, if you take $X=[0,1]$, take $\mathcal T_1$ to be the Euclidean topology and take $\mathcal T_2$ to be the topology such that the open sets are exactly $\{[0,1],\{1\},\emptyset\}$, then every net is $\mathcal T_2$-convergent to every point in $[0,1)$ (so, in particular, every $\mathcal T_1$-convergent net is trivially $\mathcal T_2$-convergent). But $\{1\}\in \mathcal T_2\setminus \mathcal T_1$.

The answer is also no if $\mathcal T_2$ is just $T_0$: if you take $X=\{0,1\}$, $\mathcal T_1=\{X,\emptyset,\{0\}\}$, $\mathcal T_2=\{X,\emptyset,\{1\}\}$, then again, all nets are convergent in both topologies, but there is no containment.

I am not sure whether there is a counterexample where $\mathcal T_1$ is Hausdorff (or even $T_1$) and $\mathcal T_2$ is $T_0$.

Answer 2

Yes, if the limits of the net are also the same for the tow topologies.

If $A \in \mathcal T_2$ the $A^{c}$ is closed in $(X, \mathcal T_2$) so the limit of any any convergent net in $A^{c}$ belongs to $A^{c}$. The same thing holds for $(X,\mathcal T_1)$ by hypothesis. Hence $A^{c}$ is closed and $A$ is open in $(X,\mathcal T_1)$.

Hint for the general case: let $(x_i)_{i \in I} \to x$ in $\mathcal T_1$. Define an ordering on $I \times \{0,1\}$ in an obvious way and let $y_{i,j}=x_i$ if $j=0$ and $x$ if $j=1$. Then $(y_{i,j})$ is convergent net in $\mathcal T_1$ hence also convergent in $\mathcal T_2$. This implies that $x_i \to x$ in $\mathcal T_2$. I am assuming that $\mathcal T_2$ is Hausdorff.

Answer 3

Yes, it holds if we include the limit points in the condition (or consider Haussdorf spaces):

Suppose that every net $(x_v)$ which converges to $a$ according to $\mathcal T_1$ also converges to $a$ according to $\mathcal T_2$.

Let $C$ be a $\mathcal T_2$-closed set, and $(x_v)$ a net converging to $a$ according to $\mathcal T_1$. Then $(x_v)$ converges to $a$ according to $\mathcal T_2$ as well, so $a\in C$ because of closedness, and that proves that $C$ is also $\mathcal T_1$-closed.

Without the extra condition, I guess, we can construct a (necessary non-Haussdorf) counterexample.