# Borel $\sigma$-algebra of a Borel subset

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?

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).