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Let $\tau_1\subseteq \tau_2$ be two Hausdorff regular topologies in an infinite set $X$ such that the convergences of sequences in $\tau_1$ and $\tau_2$ coincide (they have the same convergent sequences (to the same limits)). Must the Borel $\sigma$-fields (generated by open sets) of $\tau_1$ and $\tau_2$ be the same?

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    Of course not. Why not think about how to find a counterexample?2011-10-03
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    @GEdgar: I don't doubt you. What's wrong with my answer?2011-10-03
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    @GEdgar: sorry, I deleted my answer again because it was getting down-voted with no comments.2011-10-03
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    @robjohn: I was in the middle of typing a comment (with a counterexample) but you deleted your answer before I could post it.2011-10-03
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    @GEdgar: Is your comment "Of course not" responding to the original question, or a since-deleted comment?2011-10-03
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    @Nate:Suppose there is a set $K$ that is closed in $\tau_2$ that is not closed in $\tau_1$. Let $\bar{K}$ be the closure of $K$ in $\tau_1$. Pick a point $k\in\bar{K}\setminus\!K$. There is a sequence $k_i\in K$ converging to $k$. Since $k_i\to k$ in $\tau_1$, $k_i\to k$ in $\tau_2$. K is closed in $\tau_2$, so $k\in K$. Contradiction. What am I missing? I thought that the Borel $\sigma$-fields would be the same if the open sets were the same. Or am I misunderstanding $T3$?2011-10-03
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    I'm not 100% sure, but it seems that [descriptive-set-theory] fits here nicely.2011-10-03
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    @robjohn: (lack of) first countability.2011-10-03
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    @robjohn: If $\tau_2$ is not first countable, there can be a point in $\bar{K}$ that is not the limit of any sequence from $K$.2011-10-03
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    @Nate: and Mark, thanks. I need to review my classifications before I open my mouth again.2011-10-03

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