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

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?

• It seems you want to use $[x, x+1)=\{x\}$ which holds only in $\Bbb Z$ but not in $\Bbb R$. – Berci Aug 8 '19 at 9:52
• @Berci The $1$ in $x+1$ could be replaced by any fixed $\delta > 0$. I just need a semiopen rectangle $[x,x+\delta)×[−x,−x+\delta)$ (a $\tau_S$-open set) which intersects the antidiagonal $Y$ in exactly one point, namely the corner $(x,-x)$. This is true for any $x \in \mathbb{R}$. – yada Aug 8 '19 at 12:08
• Ah ok, I see. I misread the corners of the semiopen rectangle.. – Berci Aug 8 '19 at 12:12

I have migrated this question to mathoverflow: the claim that $$\sigma(\tau_e) = \sigma(\tau_S)$$ is not true. From $$\tau_e \subseteq \tau_S$$ we get $$\sigma(\tau_e) \subseteq \sigma(\tau_S)$$. The sets $$[a,b) \times [c,d)$$ form a base of the Sorgenfrey plane topology and this base is contained in $$\sigma(\tau_e)$$. Hence the $$\sigma$$-algebra generated by these base elements is contained in $$\sigma(\tau_e)$$. But this is different from the $$\sigma$$-algbera $$\sigma(\tau_S)$$ generated by all the open sets of the Sorgenfrey plane (which are uncountable unions of the base elements). Hence $$\sigma(\tau_e) \subsetneq \sigma(\tau_S)$$.