Proof of Subspace Topology Problem: Error Identification & Explanation

[Norashii]

Homework Statement

Let $K$ be a subset of a metric space $M$, then if $U$ is open, $U\cap K$ is open in $K$

Relevant Equations

Closed set: The set is called closed if all convergent sequences of elements of the set converge to an element of the set.

Open set: A set $X$ is said to be open if for every point $x$ in the set there is an open ball centered in it that is contained in the set.

Closure: The closure of a set $A$ is the intersection of all closed sets that contain $A$

I have already seen proofs of this problem, but none of them match the one I did, therefore I would be glad if someone could indicate where is the mistake here. Thanks in advance.**My proof:** Take a limit point x of U that is not in U, but is in K (in other words x \in K \cap(\overline{U}-U)), then suppose that K\cap U is closed, this implies \overline{K \cap U}=K \cap U and then must contain all limit points of K\cap U since x is a limit point of U and is in K, it is also a limit point of K\cap U and therefore must be in it since its closed. However, this is absurd since it would imply that x \in U then K \cap U must be open.

 

[andrewkirk]

You appear to have assumed $K\cap U$ is closed and tried to derive a contradiction from that. If you have successfully done that (I didn't check) then you can conclude that $K\cap U$ is not closed. However that does not imply that it is open. Many sets are neither open nor closed, eg the half-open interval [0,1).

Also, you have not at any point in your proof attempt referrred to being open in K, which is what you have to prove. Being open in K is different from just being open (ie open in M). In a problem like this you need to make clear, when you refer to a set being open or closed, whether you mean open or closed in K or in M.

 

[nucl34rgg]

If you look at the definition of subspace topology, and what the open sets are specifically of K, it will be immediate why the statement holds.

 

[Office_Shredder]

If you look at the definition of subspace topology, and what the open sets are specifically of K, it will be immediate why the statement holds.

I think this is probably not true. In particular K is a metric space, and so has a topology from that, which you are probably supposed to use. I suspect this question is leading up to motivating why the definition of the subspace topology makes sense.