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

  1. $\mathcal{B}_{\overline{\mathbb{R}}} = \sigma(\mathcal{T}_{\overline{\mathbb{R}}})$, that is, $\mathcal{B}_{\overline{\mathbb{R}}}$ is generated by $\mathcal{T}_{\overline{\mathbb{R}}}$;
  2. 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:

  1. Is what I'm doing correct?
  2. 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:

  1. $(a; b)$,
  2. $[x_{\min}; b)$ (provided that $x_{\min}$ exists),
  3. $(a; x_{\max}]$ (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.


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

  1. $A \cap \mathbb{R}$ is open in the standard topology of $\mathbb{R}$;

  2. if $-\infty \in A$ then there is $r\in \mathbb{R}$ such that $[-\infty, r]\subseteq A$, and

  3. 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:

Let $(X,\tau)$ be any topological space. If $f: X \to \mathbb{R}$ is continuous w.r.t. $\tau$ and the standard topology of $\mathbb{R}$ and we extend the counterdomain of $f$ to $\overline{\mathbb{R}}$ then the function $\overline{f}: X \to \overline{\mathbb{R}}$ is continuous w.r.t. $\tau$ and $\mathcal{T}_{\overline{\mathbb{R}}}$.

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 $\overline{f}=i \circ f$.

  • $\begingroup$ I think my topology $\mathcal{T}_1$ is equal to your topology $\mathcal{T}_2$. $\mathcal{T}_1 \subset \mathcal{T}_2$ since every basis element of $\mathcal{T}_1$ is open in $\mathcal{T}_2$. Now let $U \in \mathcal{T}_2$. Suppose $\infty \in U$, $-\infty \not\in U$ (other cases are analogous). We can write $$U = (s; \infty] \cup \bigcup_{x \in U \cap \mathbb{R}} (a_x, b_x),$$ which is open in $\mathcal{T}_1$. $\endgroup$ – edubrovskiy Jun 10 '16 at 10:09
  • $\begingroup$ However, I see that $\mathcal{T}_1$ and $\mathcal{T}_2$ are induced by $d$. I suppose my approach is pretty much the same as metric space approach. I just didn't like the idea of using $\arctan$ to justify arguments in measure theory, because currently I'm pretending that I don't know any calculus and want to build it with measure theory. $\endgroup$ – edubrovskiy Jun 10 '16 at 11:12
  • $\begingroup$ @edubrovskiy The $\arctan$ is only a way to create a bijection from $\overline{\mathbb{R}}$ onto $[-\frac{\pi}{2},\frac{\pi}{2}]$, and then easily transport the topology (and metric) of $[-\frac{\pi}{2},\frac{\pi}{2}]$ to $\overline{\mathbb{R}}$, so turning the bijection into an homeomorphism ( and into an isometry). $\endgroup$ – Ramiro Jun 11 '16 at 2:11

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