-1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ To check Hausdorffness use definition. $\endgroup$ – Offlaw Nov 22 '18 at 13:47
1
$\begingroup$

$U$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.