Open Subsets of open sets

Question

How does one go about proving/disproving that given $(X,d)$ a metric space that a subset $S$ is open.

Given the following definitions:

A set $X$ is open $\iff \forall x \in X, x\in int(X)$ i.e. x is interior point

where an interior point is defined as:

$x$ is a interior point of $A \subset X$ if $\exists r>0, B(z,r)\subset A \land x \in B(z,r) $

I apologise if I am being to vague and/or general

Prove that for every $s\in S$ an $r>0$ can be found such that $B(s,r)\subset S$. That proves that $S$ is open. — drhab, Mar 24 '14 at 10:49

score 1 · Accepted Answer · answered Mar 24 '14 at 10:50

1

Usually, you can prove a set is open by definition.

For example, in $\mathbb R^2$ with standard metric, the square $S=\{(x,y)| |x|<1, |y|<1\}$ is open. Why? Take any $(x,y)\in S$. Then $|x|<1$, therefore there exists some $\delta_x$ such that $(x-\delta_x, x+\delta_x)\subseteq(-1,1)$. In the same way, there exists some $\delta_y$ such that $(y-\delta_y, y+\delta_y)\subseteq(-1,1)$.

It is then easy to show that for $\delta=\min\{\delta_x,\delta_y\}$, the ball $B((x,y),\delta)$ is a subset of $S$.

answered Mar 24 '14 at 10:50

5xum

123,496
6
128
204

So the set ${x \in \Bbb R | |x| > 2}$ would be open? – Mathman Mar 24 '14 at 10:55
I don't know, would it? Can you find a ball for every $x$ in the set, contained in the set? – 5xum Mar 24 '14 at 10:59
I think so. Would it not be $(-\infty , -2) \cup (2 ,\infty)$ – Mathman Mar 24 '14 at 11:05
Indeed. And it is very easy to show that a union of open sets is always an open set. Well done. – 5xum Mar 24 '14 at 11:47

Open Subsets of open sets

1 Answers1

Linked