# Closure of Real Interval is Closed Real Interval/Proof 1

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## Theorem

Let $I$ be a non-empty real interval such that one of these holds:

- $I = \left({a \,.\,.\, b}\right)$
- $I = \left[{a \,.\,.\, b}\right)$
- $I = \left({a \,.\,.\, b}\right]$
- $I = \left[{a \,.\,.\, b}\right]$

Let $I^-$ denote the closure of $I$.

Then $I^-$ is the closed real interval $\left[{a \,.\,.\, b}\right]$.

## Proof

There are four cases to cover:

- $(1): \quad$ Let $I = \left({a \,.\,.\, b}\right)$.

From Closure of Open Real Interval is Closed Real Interval:

- $I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

- $(2): \quad$ Let $I = \left[{a \,.\,.\, b}\right)$.

From Closure of Half-Open Real Interval is Closed Real Interval:

- $I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

- $(3): \quad$ Let $I = \left({a \,.\,.\, b}\right]$.

From Closure of Half-Open Real Interval is Closed Real Interval:

- $I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

- $(4): \quad$ Let $I = \left[{a \,.\,.\, b}\right]$.

From Closed Real Interval is Closed in Real Number Line:

- $I$ is closed in $\R$.

From Set is Closed iff Equals Topological Closure:

- $I^- = \left[{a \,.\,.\, b}\right]$

$\Box$

Thus all cases are covered.

The result follows by Proof by Cases.

$\blacksquare$