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- Homework Statement
- Let [itex]K[/itex] be a subset of a metric space [itex]M[/itex], then if [itex]U[/itex] is open, [itex]U\cap K[/itex] is open in [itex]K[/itex]

- 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 [itex]X[/itex] is said to be open if for every point [itex]x[/itex] in the set there is an open ball centered in it that is contained in the set.

Closure: The closure of a set [itex]A[/itex] is the intersection of all closed sets that contain [itex]A[/itex]

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 [itex]x[/itex] of [itex]U[/itex] that is not in [itex]U[/itex], but is in [itex]K[/itex] (in other words [itex]x \in K \cap(\overline{U}-U)[/itex]), then suppose that [itex]K\cap U[/itex] is closed, this implies [itex]\overline{K \cap U}=K \cap U[/itex] and then must contain all limit points of [itex]K\cap U[/itex] since [itex]x[/itex] is a limit point of [itex]U[/itex] and is in [itex]K[/itex], it is also a limit point of [itex]K\cap U[/itex] and therefore must be in it since its closed. However, this is absurd since it would imply that [itex]x \in U[/itex] then [itex]K \cap U[/itex] must be open.