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An open subset of a metric space is a set that contains only interior points.

In the text below, $(X, d)$ will always refer to a metric space.

Definition: neighbourhood of a point

Let $x \in X$. By a neighbourhood of $x$ of radius $0 < r \in \mathbf R$ we mean the set $N_r(x) = \{ y \in X | d(x, y) < r \}$.

Definition: interior point

Let $E$ be a nonempty subset of $X$. A point $x \in E$ is said to be an interior point of $E$ if and only if there exists a neighbourhood $N$ of $x$ such that $N \subset E$.

Definition: open set

A nonempty subset $E$ of $X$ is said to be open if and only if every point of $E$ is an interior point of $E$.

The property of being open is related to the property of being closed by the following theorem.

Theorem: relation between open and closed sets

A subset $E$ of $X$ is open if and only if its complement $E^c$ is a closed subset of $X$.

Proof. First suppose that $E^c$ is closed. We want to show that this implies that $E$ is open. Choose $x \in E$. Then $x$ is not a limit point of $E^c$ (if it was, then $x$ would be an element of $E^c$, by definition of being closed, which is absurd, since $x \in E$), and hence there exists a neighbourhood $N$ of $x$ such that $N \cap E^c$ is empty. But then $N \subset E$ so that $x$ is an interior point of $E$. Hence $E$ is open.

Now suppose that $E$ is open. We want to show that this implies that $E^c$ is closed. Let $x$ be a limit point of $E^c$. If no such $x$ exists, then $E^c$ contains all its limit points, and the proof is complete. If not, then every neighbourhood $N$ of $x$ is such that $N \cap E^c$ is not empty. But then $x$ is not an interior point of $E$. Since $E$ is open, we must have $x \in E^c$. But then $E^c$ is closed, and the proof is complete.

QED.

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