Closure of Real Interval is Closed Real Interval/Proof 1
Jump to navigation
Jump to search
Theorem
Let $I$ be a non-empty real interval such that one of these holds:
- $I = \openint a b$
- $I = \hointr a b$
- $I = \hointl a b$
- $I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$.
Proof
There are four cases to cover:
- $(1): \quad$ Let $I = \openint a b$.
From Closure of Open Real Interval is Closed Real Interval:
- $I^- = \closedint a b$
$\Box$
- $(2): \quad$ Let $I = \hointr a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
- $I^- = \closedint a b$
$\Box$
- $(3): \quad$ Let $I = \hointl a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
- $I^- = \closedint a b$
$\Box$
- $(4): \quad$ Let $I = \closedint a b$.
From Closed Real Interval is Closed in Real Number Line:
- $I$ is closed in $\R$.
From Set is Closed iff Equals Topological Closure:
- $I^- = \closedint a b$
$\Box$
Thus all cases are covered.
The result follows by Proof by Cases.
$\blacksquare$