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I had a similar question as the one in Topology on the extended real line, where the answer doesn't seem very clear to me. After thinking about it, I came up with the following proof, but wasn't sure if it's correct. I'd appreciate it if someone can confirm if it's valid or point out where I'm mistaken.

Problem: Let $\tau$ be the collection of sets of the form $(a,b), [-\infty,a), (a,\infty],$ and any union of segments of this type. ($a, b\in \mathbb R$ are arbitrary real numbers.) Show that $\tau$ is a topology on the extended real line, $\overline{\mathbb R}$.

My Attempted proof: Suppose $V_\alpha\in \tau$. We need to show
(a) $\emptyset \in \tau$ and $\overline{\mathbb R}\in \tau$
(b) $\bigcup_\alpha V_\alpha\in \tau$ (arbitrary union)
(c) $V_1\cap V_2 \cap \cdots \cap V_n\in \tau$ (finite intersection)

(a) is easy, since $(1, 1)=\emptyset$ and $[-\infty, 1)\cup(-1, \infty]=\overline{\mathbb R}.$

(b) is also easy, since an arbitrary union of arbitrary unions of segments of the above type (i.e. $(a,b), [-\infty,a), (a,\infty]$) is still an arbitrary union of segments of this type.

To prove (c), it suffices to show $V_1\cap V_2\in \tau$ and then resorts to induction. Suppose $V_1=\bigcup_\alpha A_\alpha$ and $V_2=\bigcup_\beta B_\beta$, where $A_\alpha, B_\beta$ are segments of the form $(a,b), [-\infty,a),$ or $(a,\infty]$. Then $V_1 \cap V_2=(\bigcup_\alpha A_\alpha)\cap (\bigcup_\beta B_\beta)=\bigcup_\alpha\bigcup_\beta (A_\alpha \cap B_\beta)\in \tau$, since $A_\alpha \cap B_\beta$ is itself a segment of the same form.

Is this proof correct? Or am I missing something? Thanks a lot!

syeh_106
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1 Answers1

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This topology is just the description of the order topology on $\overline{\mathbb{R}}$, with its standard order where we have the usual order on $\mathbb{R}$ and $-\infty < x < +\infty$ for all $x \in \mathbb{R}$.

This is a linearly ordered set and its order topology is thus given by the subbase $\mathcal{S} = \{L_x, R_x: x \in \overline{\mathbb{R}}\}$, where $L_x = \{ y \in \overline{\mathbb{R}}: y < x\}$ and $R_x = \{y \in \overline{\mathbb{R}}: y > x\}$.

In your notation, this subbase generates the base as you describe with $(a,b) = L_b \cap R_a$, $[-\infty, a) = L_a$ and $(a, +\infty] = R_a$, and this always holds in ordered spaces (see Munkres, e.g.). Finite intersections always reduce to the intersection of a right and a left set, or just one left or right set.

Henno Brandsma
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