This is a question from math.stackexchange, which was not answered for a month now. I don't feel comfortable to post it on mathoverflow, but I am somehow blind to see the mistake in the argumentations below.

Let $(X, \tau)$ be a topological space. Then $\sigma(\tau)$ is the Borel $\sigma$-algebra on $(X, \tau)$. For any subset $Y \subseteq X$ the subspace topology on $Y$ is $\tau|Y = \{ G \cap Y \mid G \in \tau \}$ and the trace $\sigma$-algebra on $Y$ is $\sigma(\tau)|Y = \{ B \cap Y \mid B \in \sigma(\tau) \}$. It holds $\sigma(\tau|Y) = \sigma(\tau)|Y$. If $Y \in \sigma(\tau)$ then $\sigma(\tau)|Y \subseteq \sigma(\tau)$, hence $\sigma(\tau|Y) \subseteq \sigma(\tau)$.

Consider $X = \mathbb{R}^2$, $\tau_e$ the Euclidean topology and $\tau_S$ the Sorgenfrey plane topology (generated by semi-open rectangles $[a, b) \times [c, d)$). Then

  • $\tau_e \subsetneq \tau_S$ (open rectangles $(a,b) \times (c,d)$ can be written as a union of semi-open rectangles)
  • but $\sigma(\tau_e) = \sigma(\tau_S)$ (since $[a, b) \times [c, d) \in \sigma(\tau_e)$).

Consider the antidiagonal $Y := \{ (x, -x) \mid x \in \mathbb{R} \}$. Then $Y$ is a $\tau_e$-closed subset of $X$, hence a $\tau_S$-closed subset. For any $x \in \mathbb{R}$ it holds $\{ (x, -x) \} = ([x, x+1) \times [-x,-x+1)) \cap Y \in \tau_S|Y$, i.e. every point in $Y$ is $\tau_S|Y$-open in $Y$. Therefore, $\tau_S|Y = \mathcal{P}(Y)$ is the discrete topology, hence $\sigma(\tau_S|Y) = \mathcal{P}(Y)$.

Now, since $Y$ is $\tau_S$-closed in $X$, we have $Y \in \sigma(\tau_S)$ and therefore $\sigma(\tau_S|Y) \subseteq \sigma(\tau_S) = \sigma(\tau_e)$, hence $\mathcal{P}(Y) \subseteq \sigma(\tau_e)$. But this is a contradiction (e.g. by comparing the cardinalities: $|Y| = \frak{c}$, hence $|\mathcal{P}(Y)| = 2^{\frak{c}}$ while $|\sigma(\tau_e)| = \frak{c}$ because $\sigma(\tau_e)$ is generated by countably many sets (the open rectangles with rational endpoints); see also here).

What am I missing?


1 Answer 1


The problem is that you have to take uncountable unions of sets of the form $[a,b) \times [c,d)$ to get every open set in the Sorgenfrey plane, so the $\sigma$-algebra generated by $[a,b) \times [c,d)$ is strictly smaller than the Borel $\sigma$-algebra.

Which is to say, in your notation, $\sigma(\tau_e) \subseteq \sigma(\tau_S)$, but $\sigma(\tau_S) \not\subseteq \sigma(\tau_e)$.

Interestingly, it is the case that the Borel $\sigma$-algebra of the Sorgenfrey line agrees with the Borel $\sigma$-algebra of the usual topology on $\mathbb{R}$, and it is easy to give a false proof of this. The correct proof uses the hereditary Lindelöfness of the Sorgenfrey line (something not true of the Sorgenfrey plane).


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