# Given set U is first countable or not?

In $$\mathbb{R}$$ with usual topology ,the set $$U =\{ x \in \mathbb{R} : -1\le x \le 1 , ,x \neq 0\}$$ is

Choose the correct statement

$$a)$$ Neither hausdorff nor First counatble

$$b)$$ Hausdorff

$$c)$$ First countable

$$d)$$both hausdorff and first countable

My attempt :set $$U$$ can be written as $$[-1,0)$$ and $$(0,1]$$ which are two disjoint set, From this i can concnclude that $$U$$ is hausdorff

Im confusing that it is First countable or not ?

Any hints/solution will be appreciated

thanks u

## 3 Answers

The usual topology is induced by a metric and every metric space is first-countable.

$$\mathbb{R}$$ with usual topology is also a metric space. So $$\mathbb{R}$$ is first countable. Hence any subspace is also first countable.

• To check Hausdorffness use definition. – Offlaw Nov 22 '18 at 13:47

$$U$$ can be written as $$[-1,0)$$ and $$(0,1]$$ which are two disjoint set, From this i can concnclude that $$U$$ is hausdorff

The fact that $$U$$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.

For the first countable property, google is your friend.