Why if $U$ is open in $\overline {\Bbb R}$ then $U\cap \Bbb R$ is open in $\Bbb R$ in this proof?

Question

Consider $f:\mathbb R\to\overline{\mathbb R}$ defined as $f(x)=\begin{cases}\frac{1}{x}, x\neq 0\\ \infty, x=0 \end{cases}$

$f$ is Borel-measurable:

Proof (of Daniel Wainfleet).

It suffices to prove that if $U$ is open in $\overline {\Bbb R}$ then $f^{-1}U$ is Borel in $\Bbb R.$

Let $g$ be the restriction of $f$ to the domain $D=\Bbb R$ \ $\{0\}.$ Now $D$ is open in $\Bbb R$ and $g:D\to \Bbb R$ is continuous.

$\color{blue}{\text{If $U$ is open in $\overline {\Bbb R}$ then $U\cap \Bbb R$ is open in $\Bbb R$}}$ and $$f^{-1}U=A\cup B$$ where $$A=f^{-1}(U\cap \Bbb R)=g^{-1}(U\cap \Bbb R)$$ and $$B= f^{-1}(U \setminus \Bbb R).$$ Now $A$ is open in $D$ by continuity of $g,$ and $D$ is open in $\Bbb R,$ so $A$ is open in $\Bbb R.$ So $A$ is Borel in $\Bbb R.$

And $B=\emptyset$ or $B=\{0\}$ so $B$ is Borel in $\Bbb R.$

So $f^{-1}U=A\cup B$ is Borel in $\Bbb R.$

Why the blue part of the proof?

What result is being used?

Answer 1

$\color{blue}{\text{If $U$ is open in $\overline {\Bbb R}$ then $U\cap \Bbb R$ is open in $\Bbb R$}}$ it is because of the topology, this:

$$\tau_{A}=\{A\cap G: G~\text{is open in }X\}$$

when$$A\subset X$$

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In the particular case of the exercise

As $\mathbb R\subset\mathbb{\overline {R}},$ the topology of $\mathbb R$ is defined as

$\tau_{\mathbb R}=\{\mathbb R\cap U: U~\text{is open in }\overline{\mathbb R}\}$

Answer 2

$U$ is either

1. a complement of a compact set in the $\mathbb{R}$ topology, or
2. open in $\mathbb{R}$ topology.

Case 2 is trivial. For case 1, note that $U = U' \cup \{\infty\}$ so, $U\cap \mathbb{R} = U'$ and $U'$ is the $\mathbb{R}$-complement of a compact, so closed, set. So $U'$ is open and we are done.

Answer 3

$\mathbb{R}$ is a topological subspace of its two-point compactification $\overline{\mathbb{R}}$. That's is what's being used.