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I have a group $G$ with two topologies $\tau, \tau'$ on $G$ that makes it a Polish group (a completely metrizable and separable topological group). I need to show that if $\mathcal{B}(\tau)=\mathcal{B}(\tau') $ , that is they define the same borel sets, then we have $\tau=\tau'$.

I just tried the obvious, I suppose a $U\in \tau$ then $U\in \mathcal{B}(\tau)=\mathcal{B}(\tau')$ but I can't go anywhere from here...

Could you help me?

Marcelo
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1 Answers1

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It suffices to assume that they have the same sets with the Baire property.

It is enough to show that if $e\in U$ is $\tau$-open, then it contains a $\tau'$-open $V\ni e$.

Since $G$ is a $\tau$-topological group, there is a $\tau$-open $W\ni e$ with $W=W^{-1}$ and $WW\subseteq U$. But $W$ is $\tau'$-Borel, so $W$ has the Baire property with respect to $\tau'$, and it is not $\tau'$-meagre (since countably many translates cover $G$), so we can find a nonempty $\tau'$-open $V_0$ such that $W\cap V_0$ is comeagre in $V_0$, and we may assume without loss of generality that $V_0$ is symmetric (replacing it with $V_0\cup V_0^{-1}$ if necessary).

Then you can show that $V_0V_0\subseteq WW\subseteq U$. Since $V_0$ is symmetric, $e\in V_0V_0$ and we are done. Source: Automatic continuity of group homomorphisms.

tomasz
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    I think this is still your argument but I made it shorter: Consider the identity function $id:(G,\tau)\to(G,\tau')$, which is Borel and thus automatically continuous. By definition of continuity, $\tau'\subseteq\tau$. Use symmetry to see $\tau=\tau'$. –  Jan 28 '21 at 20:18
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    @EdwardH: Yes, this is the same thing, I just spelled out the proof in this particular case (with some minor simplifications). I did not want to simply apply the result, since that would not be very enlightening for OP, I think. – tomasz Jan 29 '21 at 08:39
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    I say "particular case", but I'd wager it wouldn't be too hard to show the reverse implication. – tomasz Jan 29 '21 at 08:41
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    Ah I see, thanks for clarifying! –  Jan 29 '21 at 12:39