Subspace of Complete Metric Space is Closed iff Complete
Theorem
Let $\struct {M, d}$ be a complete metric space.
Let $\struct {S, d}$ be a subspace of $\struct {M, d}$.
Then $S$ is closed if and only if $S$ is complete.
Proof
This will be proved by demonstrating the contrapositive:
- $S$ is not complete if and only if $S$ is not closed.
Necessary Condition
Suppose that $S$ is not complete.
Then there exists a Cauchy sequence $\sequence {x_n}$ in $S$ such that the limit $\ds x = \lim_{n \mathop \to \infty} x_n$, which exists in the complete metric space $M$, is not a member of $S$.
For all $\epsilon > 0$, there exists an $N \in \N$ such that for all $n \ge N$:
- $\map d {x, x_n} < \epsilon$
Hence $M \setminus S$ is not open.
Therefore, $S$ is not closed.
$\Box$
Sufficient Condition
Suppose that $S$ is not closed.
Then $M \setminus S$ is not open.
Therefore, there exists a $x \in M \setminus S$ such that for all $\epsilon > 0$, there exists a $y \in S$ such that $\map d {x, y} < \epsilon$.
So there exists a sequence $\sequence {y_n}$ in $S$ such that for all $n \in \N$:
- $\map d {x, y_n} < \dfrac 1 n$
Now, we show that $\sequence {y_n}$ is a Cauchy sequence.
Let $N \in \N$ be such that for all $n \ge N$:
- $\map d {x, y_n} < \dfrac \epsilon 2$
Let $m, n \ge N$.
Then, by Metric Space Axiom $(\text M 2)$: Triangle Inequality:
- $\map d {y_m, y_n} \le \map d {x, y_m} + \map d {x, y_n} < \epsilon$
Hence $\sequence {y_n}$ is a Cauchy sequence.
Because $\struct {M, d}$ is a complete metric space by assumption, the limit $\ds \lim_{n \mathop \to \infty} y_n$ exists and is in $M$.
Denote this limit by $y$.
By the definition of $\sequence {y_n}$:
- $\ds \lim_{n \mathop \to \infty} \map d {x, y_n} = 0$
From Distance Function of Metric Space is Continuous and Composite of Continuous Mappings is Continuous:
- $\map d {x, y} = 0$
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(we are composing the distance function with the map $y\mapsto(x,y)$.)
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By definition of a metric, this implies that $x = y$.
Since $y \notin S$, $S$ is not complete.
$\blacksquare$