## anonymous 4 years ago Is $$(0,1)\cup\{2\}$$ open or closed?

1. anonymous

Howcome? :(

2. anonymous

i would not say open because the complement is not closed

3. anonymous

This is what I did:

4. anonymous

open it is

5. anonymous

depends on your definition of open but one definition is for every element in A there is an open neighborhood contained in A that is not the case for the number 2

6. anonymous

A is open if $\forall x \in A \exists \epsilon\text { such that } (x-\epsilon,x+\epsilon)\in A$

7. anonymous

also clearly not closed, since it does not contain all its limit points (namely 0 and 1)

8. anonymous

Theorem: If a set $$A$$ has an isolated point, then it cannot be an open set. Proof: Let $$a\in A$$ be an isolated point in A. Then there exists an $$\epsilon$$-neighborhood $$V_{\epsilon}(a)$$ such that $$V_{\epsilon}(a)\cap A=\{a\}$$. It follows that $$A$$ cannot be open since there exists no $$\epsilon$$-neighborhood $$V_{\epsilon}(a)$$ such that $$V_{\epsilon}(a)\subseteq A$$. $$\blacksquare$$

9. anonymous

(with $$\epsilon>0$$, that is)