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Let $Y \subset X$ be a subset of the metric space $(X,d)$ and let $d_y$ be the restriction to $Y \times Y$. Show the following:

A subset $U \subset Y$ is open in $Y$ if and only if there exists an open subset $V \subset X$ so that $U = V \cap Y$.

First implication:

Assume $U \subset Y$ is an open subset in $Y$. We need to show that there exists an open subset $V \subset X$ so that $U = V \cap Y$.

I have absolutely no idea on how to approach this question. Can't we just say $U=V$? Since $U$ is open $V$ is open and since $U=V \subset Y$ we have $V \cap Y = V = U$?

Some help would be much appreciated!

Lorkus
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  • Your 1st direction is correct. Since every $x\in U$ is in X then it is enough to take V=U. The other direction is from the definition of the restriction. – JonesY Sep 25 '16 at 12:31

1 Answers1

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First it is important to know that being open in $Y$ is different that being open in $X$. For example, if $X=(0,1)$ the open interval, and $Y=\{\frac{1}{2}\}$ the singleton then $\{\frac{1}{2} \}$ is open in $Y$ because all of the balls in $Y$ consist of only the singleton, but $\{\frac{1}{2}\}$ is not open in $X$.

Now let us turn our attention to the problem at hand. We know from the definition of open that for $a\in U$ there exists $\epsilon_a >0$ such that $B_{d_y}(a, \epsilon_a)\cap Y \subset U$, where $B_{d_y}(a, \epsilon_a)$ is the ball of radius $\epsilon_a$ around point $a$. We also know that $B_{d_x}(a,\epsilon_a)\cap Y \subset U$ because $d_y$ is the restriction of $d_x$. The desired result is obtained if we take the set $V$ to be the collection of small enough balls around each point in $U$.

David B.
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