Notes - Metric Spaces MT23, Completeness

\[X = (0, 1], x_n = \frac 1 n\]

\[\text{convergent} \implies \text{Cauchy}\]

\[\text{Cauchy} \implies \text{bounded}\]

Every Cauchy sequence converges.

\[\mathbb R \text{ and } (-1, 1)\]

using e.g.

\[x \mapsto \frac{x}{\sqrt{1+x^2}\\,}\]

A subspace of a complete metric space is complete if and only if it is closed.

Let $X$ be a complete metric space where $S _ i$ are non-empty closed sets in $X$

\[S_1 \supseteq S_2 \supseteq \cdots\]

where $\text{diam}(S _ n) \to 0$, as $n \to \infty$, then

\[\bigcap^\infty_{n=1} S_n = \\{a\\}\]

$S _ i$ are non-empty closed sets, $S _ 1 \supseteq S _ 2 \supseteq \cdots$ and $\text{diam}(S _ n) \to 0$.

For each $n$, pick some $x _ n \in S _ n$. Then $(x _ n)^\infty _ {n = 1}$ is Cauchy, since given any $\varepsilon > 0$, there is some $N$ large enough that $\forall n > N$, $\text{diam}(S _ n) < \varepsilon$. If $n, m \ge N$, then since the $S _ i$s are nested, $d(x _ n, x _ m) \le \text{diam}(S _ N) < \varepsilon$.

Since $X$ complete, $x _ n \to a$ for some $a \in X$.

For each $i$, the nesting property implies that $x _ n \in S _ i$ for all $i \le n$.

Since each $S _ i$ is closed, $a \in S _ i$. Since this is true for all $i$,

\[a \in \bigcap^\infty_{i = 1} S_i\]

For uniqueness, suppose that

\[b \in \bigcap^\infty_{i = 1} S_i\]

Then $d(a, b) \le \text{diam}(S _ i)$ for all $i$.

Since $\text{diam}(S _ i) \to 0$, $d(a, b) = 0$, so $a = b$.

It is complete.

It is complete.

The uniform limit of continuous functions is continuous.

We know that the pointwise limit exists, since for fixed $x$, $f _ n(x)$ is a real number so each Cauchy sequence there will converge. Then show that this pointwise limit satisfies the required properties to be in $B(X)$ or $C(X)$.

• $C$ is bounded
• $C$ is closed, since $A \setminus B = A \cap B^{C}$ and so we have an intersection of closed subsets
• By the Heine-Borel theorem, $C$ is compact
• Compactness implies completeness.

Any closed subset of a complete metric space is complete.

Let $(f _ n)^\infty _ {n = 1}$ be a Cauchy sequence in $B(X)$. Then for each $x$, the sequence $(f _ n(x))^\infty _ {n = 1}$ is a Cauchy sequence of real numbers, since $\forall \varepsilon > 0$, $\exists N$ s.t. $\forall n, m \ge N$, $ \vert f _ n(x) - f _ m(x) \vert \le \vert \vert f _ n - f _ m \vert \vert _ \infty < \varepsilon$.

Let $f$ be the pointwise limit, ie..

\[f(x) = \lim_{n \to \infty} f_n(x)\]

We claim $f$ is a bounded function. Take $\varepsilon = 1$ in the definition of a Cauchy sequence, then $\exists N$ s.t. $\forall n, m \ge N$,

\[\sup_x |f_n(x) - f_m(x)| \ge 1\]

Then

\[|f_N(x) - f_n(x)| \le 1 \quad \forall n \ge N, \forall x \in X\]

Taking the limit as $n \to \infty$,

\[|f_N(x) - f(x)| \le 1\]

Since $f _ N$ is a bounded function, so is $f$.

Finally, we need to show that $f _ n \to f$ in the norm $ \vert \vert \cdot \vert \vert _ \infty$. Let $\varepsilon > 0$ and let $N$ be such that if $n, m \ge N$, $ \vert f _ n(x) - f _ m(x) \vert \le \varepsilon$ for all $x \in X$.

Then, for each fixed $n \ge N$ and $x \in X$, we can let $m \to \infty$, obtaining $ \vert f _ n(x) - f(x) \vert \le \varepsilon$. Then $\forall n \ge N$, $ \vert \vert f _ n - f \vert \vert _ \infty \le \varepsilon$. It follows $f _ n \to f$ in the $ \vert \vert \cdot \vert \vert _ \infty$ norm.

Since we are assuming that $B(X)$ is complete, it suffices to show that $C _ B(X)$ is a closed subset of $B(X)$ (because subsets of complete metric spaces are complete iff they are closed).

Suppose $(f _ n)^\infty _ {n = 1}$ is a sequence of elements of $C _ B(X)$ convering in the $\infty$-norm to some $f \in B(X)$. Fix some $a \in X$ for which we want to show continuity at and let $\varepsilon > 0$. Since $f _ n \to f$ in the $\infty$-norm, there $\exists n$ s.t.

\[||f_n - f||_\infty \le \varepsilon / 3\]

Since $f _ n$ is continuous, there exists $\delta > 0$ s.t.

\[|f_n(x) - f_n(a)| < \varepsilon/3\]

for all $x \in B(a, \delta)$, so we have

\[\begin{aligned} |f(x) - f(a)| &\le |f(x) - f_n(x)| + |f_n(x) - f_n(a)| + |f_n(a) - f(a)| \\\\ &< \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\\\ &= \varepsilon \end{aligned}\]

so the limit is also continuous, hence $C _ B(X)$ is a closed subset of $C(X)$, so it is complete.

Proof sketch:

• If a subset $Y$ is closed in $X$, a Cauchy sequence in $Y$ is also a Cauchy sequence in $X$. Since $Y$ is closed, the limit must be in $Y$.
• If a subset $Y$ is complete, then any convergent subsequence is also convergent to an element of $Y$, so it is closed.