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I need help proving the following:
Let $(X, d)$ be a metric space and $Y\subseteq X$ be a subspace. Show that if Y is open in X and U is open in Y, then U is open in X.

I know that if Y is open in X then ($\forall$$x$$\in$ Y) ($\exists$$r$>0) B($x$,$r$)$\subset$Y. The same goes if U is open in Y. But how can I prove that U is open in X? Should I use some other definitions?

TanEma
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    $U$ open in $Y\implies U=A\cap Y$ where $A$ is open in $X$;; Also $Y$ open in $X\implies U$ open in $X$ – Learnmore Nov 06 '16 at 17:40
  • When you asked to proof "U is open in X", as in other places in math, don't try to use the hypothesis first, but try to prove the result(conclusion). And in proving the result, that is the time you may use your hypothesis. – Eric Nov 06 '16 at 17:40
  • okay guys, thanks a lot. – TanEma Nov 06 '16 at 17:42
  • To claim "U is open in X". First, choose a point $x\in U$, since $U$ is open in $Y$, then there is a ball centered at $x$ and fully contained in $Y$. At the same time $Y$ is a subset of $X$, so this ball in trivially contained in $X$, so $x$ is an interior point of $X$. Since the point $x$ is arbitrarily chosen, then for every $x$ in $U$ is an interior point. – Eric Nov 06 '16 at 17:42
  • Oh. But is that enough to prove the statement or do I need to use something else? – TanEma Nov 06 '16 at 17:45
  • My proof is valid and correct. There is no other definition or theorem to use. But for mathematical writing, one should have better write it in a more symbolic way and clear way. i.e. use the terminology like $\subseteq$, $\in$, and explicitly write down the ball like $B_r^U(x)$. I'm willing to write it down, but I've to study my exam. :) Maybe other one will help you. – Eric Nov 06 '16 at 17:49
  • Okay, thanks a lot. Good luck with your exam :) – TanEma Nov 06 '16 at 17:53

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According to definition of subspace topology, a set $A\subset Y$ is open in $Y$ if and only if $A=B\cap Y$ for some open set $B\subset X$.

So now $U$ being open in $Y$, there exists $A$ open in $X$ such that $U=A\cap Y$. Now $A,Y$ are open in $X$, hence their intersection (being finite intersection) is again open in $X$.

So, all you need to check is that, the subspace topology and the induced metric topology are same on $Y$ whenever you have $(X,d)$ a metric space. This I will leave as an exercise.

Landon Carter
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  • I think the OP's question is quite easy and trivial, but the theorem you used here are not and needs more things to say(and to prove). I personally think it is better to prove the OP's question directly by definition. – Eric Nov 06 '16 at 17:45
  • Thank you very much kind sir. – TanEma Nov 06 '16 at 17:46
  • Yes @Eric, in hindsight I realized it's better to directly prove it. :) I guess someone will provide that direct proof. So my "exercise" is heavyweight, it seems. – Landon Carter Nov 06 '16 at 17:47