# Is $\mathbb{R}$ with the particular point topology and open half line topology compact?

Consider $(\mathbb{R}, \tau_{p})$ and $(\mathbb{R}, \tau_l)$ (or sometimes called "ray topology")

Where $\tau_{p} = \{U \subseteq \mathbb{R}, p \in U\}\cup\{\varnothing\}$, and $\tau_l = \{(a, \infty)| a \in \mathbb{R}\}\cup\{\varnothing, \mathbb{R}\}$

Question: Are $(\mathbb{R}, \tau_{p})$ and $(\mathbb{R}, \tau_l)$ compact?

1. Let $\mathcal{U}$ be an open cover of $(\mathbb{R}, \tau_{p})$, then it necessarily contains $\{\mathbb{R}\}$, since $p \in \mathbb{R}$. Then we can remove all other open sets, leaving only $\{\mathbb{R}\}$. So every open cover has a finite subcover.

2. Let $\mathcal{U}$ be an open cover of $(\mathbb{R}, \tau_{l})$. Since $\mathbb{R}$ is uncountabe, to produce a finite subcover, we must remove all but finite number of sets in $\mathcal{U}$. Suppose we have removed all but finite number of sets in $\mathcal{U}$. Since $\mathbb{R}$ has no least element, therefore we can always find $a \in \mathbb{R}$ such that no cover contains it, therefore $(\mathbb{R}, \tau_{l})$ is not compact.

I know my arguments sort of sucks, I would appreciate if someone can check if these are correct and any improvement on my arguments are appreciated!

## 2 Answers

HINT:

1. Consider $\mathcal{U}=\{(-n,n): n\in\mathbb{N}\}$. Has it a finite subcover?

2. Is already done by you.

• For 1. No, but I do not know how to argue this rigorously. It is similar to the second part. – Shamisen Expert Aug 1 '16 at 6:07
• @Unicorn-SharkInvasion Any finite subcover is a set $(-n,n)$ for some $n$. – Przemysław Scherwentke Aug 1 '16 at 6:09
1. Careful! not every open cover needs to contain $\mathbb{R}$. You will have to use a different approach.

2. This is fine but actually I don't think you need uncountablility of $\mathbb{R}$ here. For instance, I think your argument would work for $\mathbb{Z}$ as well.

• Think of problem 1 as very similar to problem 2 except that in 2, $p= \infty$. Then adjust your argument for 2 to make it work for 1. – Alexander Aug 1 '16 at 14:42