Show that $\tau$ is a topology on $\mathbb{R}$

Question

I really wanted to solve this math problem on my own, but I have absolutely no idea how to attack this exercise and REALLY needs some hints:

Let $\tau$ be the system of subsets U in $\mathbb{R}$ which is one of the following types:

Either:

(i) U does not contain $0$,

(ii) U does contain $0$, and the complementary set $\mathbb{R}$ \ U is finite.

SHOW that $\tau$ is a topology on $\mathbb{R}$.

Answer 1

All you need to do is check that $$\tau = \{ U \subseteq \mathbb{R} \ | \ U \text{ doesn't contain $0$ or U contains $0$ and $\mathbb{R} \setminus U$ is finite}\} $$ actually satisfies the axioms for a topology on $\mathbb{R}$.

So you need to do the following

1. Show that $\mathbb{R}$ and $\emptyset$ are elements of $\tau$
2. Choose any collection of elements $\{U_i\}_{i \in I}$ of $\tau$ (i.e $U_i \in \tau$ for each $i \in I$) and show that $\bigcup_{i \in I} U_i \in \tau$
3. Choose any finite collection of elements of $\tau$, $\{V_1, \dots, V_n\}$ and show that $\bigcap_{i=1}^n V_i \in \tau$

Then $\tau$ is a topology on $\mathbb{R}$, by the definiton of a topology on a set.

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As an example I'll check one small part of the above for you. I'll show that $\mathbb{R} \in \tau$. Note that $\mathbb{R}$ contains $0$ and $\mathbb{R} \setminus \mathbb{R} = \emptyset$ which is certainly finite, and so $\mathbb{R} \in \tau$.