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?

  • $\begingroup$ Of course not. Why not think about how to find a counterexample? $\endgroup$ – GEdgar Oct 3 '11 at 21:22
  • $\begingroup$ @GEdgar: I don't doubt you. What's wrong with my answer? $\endgroup$ – robjohn Oct 3 '11 at 21:28
  • $\begingroup$ @GEdgar: sorry, I deleted my answer again because it was getting down-voted with no comments. $\endgroup$ – robjohn Oct 3 '11 at 21:30
  • $\begingroup$ @robjohn: I was in the middle of typing a comment (with a counterexample) but you deleted your answer before I could post it. $\endgroup$ – Nate Eldredge Oct 3 '11 at 21:32
  • 1
    $\begingroup$ I'm not 100% sure, but it seems that [descriptive-set-theory] fits here nicely. $\endgroup$ – Asaf Karagila Oct 3 '11 at 21:43

They need not be the same. Let $X = ^{\omega_1}\mathbb{Z}$, the set of integer sequences of length $\omega_1$, and give it the order topology $\tau_1$ induced by the lexicographic order. Every linearly ordered topological space is $T_1$ and hereditarily normal, so $\langle X,\tau_1 \rangle$ is certainly $T_3$. There are no non-trivial convergent sequences in $\langle X,\tau_1 \rangle$. To see this, suppose that $x \in X$, and $A \subseteq X\setminus\{x\}$ is countable. There is an $\alpha < \omega_1$ such that for each $y\in A$ there is some $\xi < \alpha$ such that $x(\xi)\ne y(\xi)$. Let $x^-,x^+\in X$ be be defined as follows: $$\begin{align*} x^-(\xi) &= \begin{cases} x(\xi),&\text{if }\xi \ne \alpha\\ x(\alpha)-1,&\text{if }\xi = \alpha \end{cases}\\ &\\&\\&\\ x^+(\xi) &= \begin{cases} x(\xi),&\text{if }\xi \ne \alpha\\ x(\alpha)+1,&\text{if }\xi = \alpha. \end{cases} \end{align*}$$

Then $x \in (x^-,x^+) \subseteq X\setminus A$.

Now let $\tau_2$ be the discrete topology on $X$; clearly $\tau_1 \subseteq \tau_2$, and there are no non-trivial convergent sequences in $\langle X,\tau_2 \rangle$, either. But the Borel $\sigma$-field of $\tau_2$ is $\wp(X)$, and I’m reasonably sure that that of $\tau_1$ isn’t.

Edit: I originally had $X = ^{\omega_1}2$ with the lexicographic order topology, which, as Byron Schmuland pointed out, does have convergent sequences. I was actually thinking of the tree topology on $X$, which has a base consisting of all sets of the form $\{x \in X:x \upharpoonright\alpha = \varphi\}$, where $\alpha < \omega_1$ and $\varphi \in ^{\alpha}2$. It’s zero-dimensional and $T_1$, hence completely regular, and every countable subset is easily seen to be closed and discrete by an argument similar to the one used above to show that $\langle X,\tau_1 \rangle$ has no non-trivial convergent sequences..

  • $\begingroup$ I am way outside my comfort zone, but I'd like to try to understand your example. Could you define "character", as in "every point has character $\omega_1$"? My naive thought is that the sequence $1_n$ converges to $1_\omega$ in the lexicographic order ($n$ is an ordinary integer, $\omega$ the first infinite ordinal, and $1_s$ means the binary sequence that is all zero, except a $1$ at $s$). Where do I misunderstand? $\endgroup$ – user940 Oct 4 '11 at 0:02
  • $\begingroup$ @Byron: The character of a point is the minimum cardinality of a local base at the point. You’re right about the sequence: I was thinking of the tree topology, not the order topology. Alternatively, you could replace $2$ by $\mathbb{Z}$ and use the lexicographic order topology. I’ll fix it. $\endgroup$ – Brian M. Scott Oct 4 '11 at 1:08
  • $\begingroup$ Why does $(1_n)_{n<\omega}$ converge to $1_\omega$ in the order topology? Isn't $(\leftarrow,s)$, where $s$ is the sequence that is zero except at $\omega$ and $\omega+1$, a neighbourhood of $1_\omega$ that does not contain any of the $1_n$'s? $\endgroup$ – LostInMath Oct 4 '11 at 1:24
  • $\begingroup$ @LostInMath: You’re right. I misread @Byron’s example as one that does work: for $n\le\omega$ let $1_n$ have $1$’s in the first $n$ places. Then $\langle 1_n:n\in\omega\rangle\to 1_\omega$. $\endgroup$ – Brian M. Scott Oct 4 '11 at 1:35
  • $\begingroup$ @Lost and Brian: Thanks for sorting that out for me! $\endgroup$ – user940 Oct 4 '11 at 1:40

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