Question

Is \((0,1)\cup\{2\}\) open or closed?

Answer 1

Howcome? :(

Answer 2

i would not say open because the complement is not closed

Answer 3

This is what I did:

Answer 4

open it is

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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 \(V_{\epsilon}(a)\subseteq A\). \(\blacksquare\)

Answer 9

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