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.

## References

- Rudin, Walter:
*Principles of Mathematical Analysis,*3rd edition, McGraw Hill, 1976.