5

Let X be a metric space and let $S\subset X$

I want to show that $U \subset S$ open in S $\iff$ there exists an open set $U'$ in $X$ such that $U=U' \cap S $

Here is a little bit of my reasonning:

For $\implies$

  • $U$ is open in $S$, we simply have to set $U = U'$ and $U = U'\cap S$

For $\Longleftarrow$

If there exists an open set $U'$ in $X$ such that $U=U'\cap S$ then either:

  • If S is closed then $U' \neq S$ but since U is an open subset of , $U' \subset S$ is also open in S. We simply have $U = U'$
  • If S is open we simply have $U' = S$ therefore $U = U'\cap S$ is open

Here is an image that might help you visualize the situation

enter image description here

My reasonning is not too rigorous, could someone help me improve it?

aribaldi
  • 1,400

2 Answers2

3

Let $B_\epsilon(x,S)$ denote the open ball centered at $x$ with radius $\epsilon$ with respect to the topology on $S$ and define $B_\epsilon(x, X)$ analogously.

($\Rightarrow$) If $U \subseteq S$ is open, then for all $x \in S$ there exists $\epsilon>0$ with $B_\epsilon(x, S) \subseteq U$. Note that $B_\epsilon(x,S)=B_\epsilon(x, X) \cap S$. Then $U= (\bigcup_{x \in S} B_\epsilon(x,X))\cap S$. The set at the left of the intersection is open in $X$ by definition.

($\Leftarrow$) Now suppose $U \subseteq S$, $U= U' \cap S$ for some open $U' \subseteq X$. Then for all $x \in U$ there exists $\epsilon>0$ such that $B_\epsilon(x, X) \subseteq U'$. Then $B_\epsilon(x,S)= B_\epsilon(x,X) \cap S \subseteq U$, i.e. since $x$ was arbitrary, $U$ is open in $S$.

M10687
  • 3,250
2

$U\subset S \implies U \subset X$

Also, $U$ open in $S$ implies that there exists a $V \subset X$ such that $V$ is open in $X$ and $U=V\cap S$ (by definition, putting subspace topology on $S$).

Conversely, let there be $U'\subset X$ such that $U=U'\cap S$, with $U'$ open in $X$. Then clealy, by the subspace topology induced on $S$ by $X$, we have $U$ to be open in $S$. So we are done.

Baby elephant
  • 315
  • 1
    I dont think you need X to be a metric space, this is true for any topological space till you are applying the subspace topology on S, which is equivalent to applying metric on S – Baby elephant Jun 06 '16 at 15:05
  • You are correct, but if you assume $S$ has the subspace topology then the statement is true by definition. I think the point is to prove it explicitly for a metric space. – M10687 Jun 06 '16 at 15:08
  • yes but topology induced by metric on S and subspace topology induced on S by metric topology on X are the same – Baby elephant Jun 06 '16 at 17:35
  • Yes, exactly, that's what the question is asking for a proof of. – M10687 Jun 06 '16 at 17:36