Subspace of Complete Metric Space is Closed iff Complete

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Theorem

Let $\left({M, d}\right)$ be a complete metric space.

Let $\left({S, d}\right)$ be a subspace of $\left({M, d}\right)$.


Then $S$ is closed iff $S$ is complete.


Proof

Necessary Condition

Suppose that $S$ is not complete.

Then there exists a Cauchy sequence $\langle {x_n} \rangle$ in $S$ such that the limit $\displaystyle x = \lim_{n \to \infty} x_n$, 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 $n \ge N$:

$d \left({x, x_n}\right) < \epsilon$

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 $d \left({x, y} \right) < \epsilon$.

So there exists a sequence $\langle {y_n} \rangle$ in $S$ such that for all $n \in \N$:

$\displaystyle d \left({x, y_n}\right) < \frac 1 n$


Now, we show that $\langle {y_n} \rangle$ is a Cauchy sequence.

Let $N \in \N$ be such that for all $n \ge N$:

$\displaystyle d \left({x, y_n}\right) < \frac \epsilon 2$

Let $m, n \ge N$.

Then, by the triangle inequality:

$d \left({y_m, y_n}\right) \le d \left({x, y_m}\right) + d \left({x, y_n}\right) < \epsilon$

Hence $\langle {y_n} \rangle$ is a Cauchy sequence.


Because $\left({M, d}\right)$ is a complete metric space by assumption, the limit $\displaystyle \lim_{n \to \infty} y_n$ exists and is in $M$.

Denote this limit by $y$.

By the definition of $\langle {y_n} \rangle$:

$\displaystyle \lim_{n \to \infty} d \left({x, y_n}\right) = 0$

Because a metric is continuous and the composition of two continuous mappings is continuous:

$d \left({x, y}\right) = 0$


By the definition of a metric, this implies that $x = y$.

Since $y \notin S$, $S$ is not complete.

$\blacksquare$