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?

  • $\begingroup$ It seems you want to use $[x, x+1)=\{x\}$ which holds only in $\Bbb Z$ but not in $\Bbb R$. $\endgroup$ – Berci Aug 8 '19 at 9:52
  • $\begingroup$ @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 $x \in \mathbb{R}$. $\endgroup$ – yada Aug 8 '19 at 12:08
  • $\begingroup$ Ah ok, I see. I misread the corners of the semiopen rectangle.. $\endgroup$ – 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 $\sigma(\tau_e) \subsetneq \sigma(\tau_S)$.

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