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Let $X\supseteq Y$ be sets, let $\tau$ be a topology on $X,$ and let $\tau_Y$ be the subspace topology on $Y;$ then $$\sigma_Y(\tau_Y)=\{A\cap Y:A\in\sigma(\tau)\},$$ where for $Z\subseteq Y$ we define $\sigma_Y(Z)$ to be the $\sigma$-algebra generated, on $Y,$ by $Z.$ (This result is proved in e.g. user642796's Math.Stackexchange post.)

However, doesn't Example 4.12 in Wise & Hall's Counterexamples in Probability and Real Analysis contradict this result? WH E4.12

(In the notation of the above example, our result says $\sigma_B(\tau_B)=\{A\cap B:A\in\sigma(\tau)\},$ so that every $C\in\sigma_B(\tau_B)$ is of the form $A\cap B$ with $A\in\sigma(\tau);$ but $B$ is closed and hence is in $\sigma(\tau),$ so $A\cap B\in\sigma(\tau),$ contradicting the statement of the Example.)

Maybe Wise & Hall made a mistake somewhere...

xFioraMstr18
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  • For the Sorgenfrey line $\Bbb S$ we do have that $\operatorname{Bor}(\Bbb S)=\operatorname{Bor}(\Bbb R)$ but this is no longer true for their squares $H$ and $\Bbb R^2$; the former has way more Borel sets, and is not the product $\sigma$-algebra of the factor spaces. I think that's the core issue. The identity for the $\sigma$-algebras of subspaces does hold, and $H$ is not a counterexample to it. – Henno Brandsma Jan 16 '21 at 08:41

1 Answers1

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There is indeed an error: The Borel $\sigma$-algebra of $H$ properly contains the Borel $\sigma$-algebra of $\Bbb R^2$. This is clear from the fact that every subset of the reverse diagonal $y=-x$ is closed in $H$ and therefore a Borel set in $H$, but not every subset of the reverse diagonal is a Borel set in $\Bbb R^2$. For further explanation see the answer to this MathOverflow question.

Brian M. Scott
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