1
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
6
$\begingroup$

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).

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.