# Topology for convergent sequences

Let $(X,\tau)$ be a topological space, and consider the family $\mathcal{F}$ of the topologies over $X$ such that the convergent sequences for each $\gamma \in \mathcal{F}$ are the same as the convergent sequences for $\tau$, with the same limits. That is, $$\mathcal{F} = \{\gamma \;|\; x_n \xrightarrow{\gamma} x \Leftrightarrow x_n \xrightarrow{\tau} x\}$$

It is easy to see that the topology $\tau_M$ generated by the topologies in $\mathcal{F}$ is such that $\tau_M \in \mathcal{F}$. The $\tau_M$ is the finest topology with the same convergent sequences as $\tau$, converging to the same point.

But is it true that the topology $$\tau_m = \bigcap_{\gamma \in \mathcal{F}} \gamma$$ is itself in $\mathcal{F}$?

In other words, does it follow that the convergent sequences for $\tau_m$ and their limits are the same as the convergent sequences and their limits in $\tau$? Are there any nice counterexamples?

In other words again, do we have a weakest topology such that the convergent sequences and their limits are the same as the ones for $\tau$?

Edit: Fixed according to @joriki's comment.

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Your first statement about $\tau_M$ doesn't hold for an indiscrete space with two elements. In $(\{1,2\},\{\{\},\{1,2\}\})$ all sequences converge (to both $1$ and $2$). This is also the case for both $(\{1,2\},\{\{\},\{1\},\{1,2\}\})$ (all sequences converge to $2$) and for $(\{1,2\},\{\{\},\{2\},\{1,2\}\})$ (all sequences converge to $1$), so here $\tau_M$ is the discrete topology, which is not in $\mathcal F$. Perhaps you want the topologies to be Hausdorff? –  joriki May 19 '13 at 2:17
Alternatively, you might want to require not only that the same sequences converge, but that they converge to the same limits. –  joriki May 19 '13 at 4:27
@joriki: You are right. That's what I meant. I will fix the post. –  André Caldas May 19 '13 at 17:31
I have a hunch it is false; no counterexample yet, though we might need infinitely many topologies for this to fail. –  Henno Brandsma May 19 '13 at 17:41
@Henno: Your hunch is right, assuming that I made no mistake. –  Brian M. Scott May 20 '13 at 6:33

Taking the intersection can introduce new convergent sequences.

Let $p$ be a point not in $\omega$. Let $X=\{p\}\cup\omega$. For each free ultrafilter $\mathscr{U}$ on $\omega$ define a topology $\tau_{\mathscr{U}}$ on $X$ as follows. Points of $\omega$ are isolated, and the basic open nbhds of $p$ in $\tau_{\mathscr{U}}$ are the sets of the form $\{p\}\cup U$ such that $U\in\mathscr{U}$. The only convergent sequences of $\langle X,\tau_{\mathscr{U}}\rangle$ are the trivial ones. Let $$\tau=\bigcap_{\mathscr{U}\in\beta\omega\setminus\omega}\tau_{\mathscr{U}}\;.$$ A set $W\subseteq X$ belongs to $\tau$ iff $p\notin W$, or $W\in\mathscr{U}$ for all $\mathscr{U}\in\beta\omega\setminus\omega$. But the only subsets of $\omega$ that belong to all free ultrafilters on $\omega$ are the cofinite sets, and It follows that

$$\tau=\wp(\omega)\cup\{W\subseteq X:p\in W\text{ and }|\omega\setminus W|<\omega\}\;.$$

That is, points of $\omega$ are isolated in $\langle X,\tau\rangle$, and the basic open nbhds of $p$ are the sets of the form $\{p\}\cup(\omega\setminus F)$ such that $F$ is finite. But then $\langle k:k\in\omega\rangle$ is a sequence in $X$ that $\tau$-converges to $p$.

Now let $\tau'$ be the intersection of all topologies on $X$ having only the trivial convergent sequences. Clearly $\tau'\subseteq\tau$, so $\langle k:k\in\omega\rangle$ converges to $p$ in $\langle X,\tau'\rangle$.

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Nice, and quite simple. So to be explicit: we take one particular $\tau_{\mathcal{U}}$ as our starting topology on $X$, and then we only have trivial convergent sequences; $\tau' = \tau_m$ gets a new convergent sequence, $\omega$ itself. –  Henno Brandsma May 20 '13 at 6:52
@Henno: Yep, in the OP’s original formulation. –  Brian M. Scott May 20 '13 at 6:54
@BrianM.Scott: Very nice! I had thought of a circle (in $\mathbb{R}^2$ plus $p = (0,0)$. But I was trying to use the topology of the circle. Taking the discrete topology on the circle, and changing the neighborhoods of $p$ would work, right? –  André Caldas May 20 '13 at 16:53
@André: It should, if you do something like what I did here to tie $p$ to the circle. –  Brian M. Scott May 20 '13 at 16:56