Set closed in Y, but not in superset of Y, X?

Homework Statement

Let Y be a subset of X. Give an example where a set A is closed in Y but not closed in X.

Homework Equations

A set is closed if its complement is open.
A set is open if for every element x0 of the set, there exists an E > 0 such that U(x0;E) = {x|d(x,x0)< E} is contained in the set.

The Attempt at a Solution

I tried various combinations of X and Y, usually leaving X as the real numbers. I started with X as the reals, and Y as some open or closed interval in the reals, but had no luck. I know that a closed set in the reals (here, X) is a set that contains its boundaries, so I am looking for a set that is missing at least one of its boundary points (and is thus non-closed in X) but is closed in Y. However, I can't seem to find a proper Y such that such a set is closed in Y. While reading through some texts online, I came across the claim that the set A = [0,1] ∩ Q of rational numbers between 0 and 1 (inclusive) is closed in the space of rational numbers, but is not closed in the real numbers. However, I can't see why this is true. The complement of A in Y is (0,1) ∩ {irrationals}, but if we choose a point y from this set, it doesn't seem like there is an E > 0 such that every point within a distance E from y is irrational.

As for A = [0,1] ∩ Q being non-closed in X, its complement there is (-∞,0)U((0,1)∩Q)U(1,∞). (0,1)∩Q is not open in the X, so neither is its union with the open intervals shown.

I just don't understand why it is closed in Y.

lanedance
Homework Helper
how about considering when $A \cap Y = Y$

clearly A is closed in Y, though you should be able to come up with a case when it i not closed in X...

If A ∩ Y = Y, the complement of A in Y is the null set, which is simultaneously open and closed, I believe. This tripped me up, but I guess that all that matters is that the complement is open for the set to be closed, correct? If that's true, then A and Y could both be the same open interval/region in R or RxR, right?

Dick