Let $\mathcal{B}_{X}$ denote the Borel $\sigma$-algebra on $X$.
I'm reading a book on real analysis by Folland and he defines
$$\mathcal{B}_{\overline{\mathbb{R}}} = \{ E \mid E \cap \mathbb{R} \in \mathcal{B}_{\mathbb{R}} \},$$
where $\overline{\mathbb{R}}$ is the extended real line. He says this agrees with the usual definition of Borel sets if we make $\overline{\mathbb{R}}$ into a metric space with metric
$$d(x, y) = |\arctan x – \arctan y|.$$
I don't like this approach because it's tricky, unintuitive and uses a metric when we only need a topology (if I'm not mistaken). I want to define a topology $\mathcal{T}_{\overline{\mathbb{R}}}$ on $\overline{\mathbb{R}}$ in such a way that
- $\mathcal{B}_{\overline{\mathbb{R}}} = \sigma(\mathcal{T}_{\overline{\mathbb{R}}})$, that is, $\mathcal{B}_{\overline{\mathbb{R}}}$ is generated by $\mathcal{T}_{\overline{\mathbb{R}}}$;
- if $f: X \to \mathbb{R}$ is continuous w.r.t. standard topology of $\mathbb{R}$ and we extend range of $f$ to $\overline{\mathbb{R}}$ then extended function $\overline{f}$ is continuous w.r.t. $\mathcal{T}_{\overline{\mathbb{R}}}$.
I have two questions:
- Is what I'm doing correct?
- Is this the standard way to handle $\pm\infty$?
Let $X$ be a chain. Then the order topology on $X$ is generated by basis consisting of sets of the following types:
- $(a; b)$,
- $[x_{\min}; b)$ (provided that $x_{\min}$ exists),
- $(a; x_{\max}]$ (provided that $x_{\max}$ exists).
Let $\mathcal{T}_{\overline{\mathbb{R}}}$ be the order topology of $\overline{\mathbb{R}}$. Choose any $U \in \mathcal{T}_{\overline{\mathbb{R}}}$ and write $U \cap \mathbb{R}$ as
$$U \cap \mathbb{R} = \bigcup_{x \in U}(B_x \cap \mathbb{R}),$$
where $B_x \subset U$ is a basis element containing $x$. This set is open in standard topology since each $B_x \cap \mathbb{R}$ is open. Thus $U \cap \mathbb{R} \in \mathcal{B}_{\mathbb{R}}$ and $U \in \mathcal{B}_{\overline{\mathbb{R}}}$. Since $U$ was arbitrary $\mathcal{B}_{\overline{\mathbb{R}}} \supset \sigma(\mathcal{T}_{\overline{\mathbb{R}}})$
It's easy to show that $\mathcal{B}_{\overline{\mathbb{R}}} = \sigma(\mathcal{E})$, where
$$\mathcal{E} = \{ (a; \infty] \mid a \in \mathbb{R} \}.$$
Now (1) follows from
$$\sigma(\mathcal{T}_{\overline{\mathbb{R}}}) \supset \sigma(\mathcal{E}) = \mathcal{B}_{\overline{\mathbb{R}}}.$$
To prove (2), suppose $f: X \to \mathbb{R}$ is continuous w.r.t. standard topology and let $U \in \mathcal{T}_{\overline{\mathbb{R}}}$. Then $\overline{f}\,^{-1}(U) = f^{-1}(U \cap \mathbb{R})$ is open since $U \cap \mathbb{R}$ is open in standard topology.
Best Answer
The topology $\mathcal{T}_{\overline{\mathbb{R}}}$ you are looking for is defined in the following way: $A\in\mathcal{T}_{\overline{\mathbb{R}}}$ iff
$A \cap \mathbb{R}$ is open in the standard topology of $\mathbb{R}$;
if $-\infty \in A$ then there is $r\in \mathbb{R}$ such that $[-\infty, r]\subseteq A$, and
if $+\infty \in A$ then there is $s\in \mathbb{R}$ such that $[s, +\infty]\subseteq A$.
It is easy to prove that $\mathcal{T}_{\overline{\mathbb{R}}}$ is a topology. In fact this is precisely the topology induced by the metric $$d(x, y) = |\arctan x - \arctan y|$$ This topology makes $\overline{\mathbb{R}}$ a compact space. It is also easy to prove that $\mathcal{B}_{\overline{\mathbb{R}}} = \sigma(\mathcal{T}_{\overline{\mathbb{R}}})$.
Now for the result:
The resul is immediate. Just note that $i: \mathbb{R} \to \overline{\mathbb{R}}$ defined by $i(x)=x$, for all $x \in \mathbb{R}$, is continuous w.r.t. the standard topology of $\mathbb{R}$ and $\mathcal{T}_{\overline{\mathbb{R}}}$ and $\overline{f}=i \circ f$.