Metric Spaces MT23, Completeness
Flashcards
Can you give an example of a metric space and a sequence such that the sequence is Cauchy but not convergent?
What implication links convergent sequences and Cauchy sequences in metric spaces?
What implication links Cauchy sequences and bounded sequences in metric spaces?
What does it mean for a metric space to be complete?
Every Cauchy sequence converges.
Can you give an example to show that completeness is not preserved under homeomorphisms?
\(\mathbb R \text{ and } (-1, 1)\) using e.g. \(x \mapsto \frac{x}{\sqrt{1+x^2}\\,}\)
What “if and only if” do we have for when a subspace of a complete metric space is also complete?
A subspace of a complete metric space is complete if and only if it is closed.
Can you state Cantor’s intersection theorem?
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 \(\bigcap^\infty _ {n=1} S _ n = \\{a\\}\)
If $X$ is a complete metric space, what conditions do you need on sets $S _ i$ for Cantor’s intersection theorem so that $\bigcap^\infty _ {n=1} S _ n = \{a\}$?
$S _ i$ are non-empty closed sets, $S _ 1 \supseteq S _ 2 \supseteq \cdots$ and $\text{diam}(S _ n) \to 0$.
Quickly prove Cantor’s intersection theorem, i.e. that if
- $X$ is a complete metric space
- $S _ i$ are non-empty closed sets
- $S _ 1 \supseteq S _ 2 \supseteq \cdots$
- $\text{diam}(S _ n) \to 0$
Then
\(\bigcap^\infty _ {i = 1} S _ i = \\{a\\}\)
for some $a \in X$.
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 $a = b$.
When $X = \mathbb R$, how can you restate the fact that $C _ b(X)$ is complete more familiarly?
The uniform limit of continuous functions is continuous.
What’s the basic idea behind proving the completness of $B(X)$ or $C(X)$ with the $\sup$ norm, given a Cauchy sequence of functions $(f _ n)$?
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 $C(X)$.
The Cantor set is defined by
\(C = [0, 1] \setminus \bigcup _ {m=1}^\infty \bigcup _ {k=0}^{3^{m-1} - 1} \left( \frac{3k+1}{3^m}, \frac{3k + 2}{3^m} \right)\)
Proving that it’s complete looks like a nightmare. What chain of deductions might you use in situations like this to actually prove this quite easily?
- $C$ 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.
What more general theorem allows you to deduce that any closed subset of $\mathbb R$ is complete?
Any closed subset of a complete metric space is complete.
Quickly prove that $B(X)$ is a complete metric space with respect to $\infty$-norm.
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. $\forall n, m \ge N$, \(\sup_x \vert f_n(x) - f_m(x) \vert \ge 1\) Then \(\vert f_N(x) - f_n(x) \vert \le 1 \quad \forall n \ge N, \forall x \in X\) Taking the limit as $n \to \infty$, \(\vert f_N(x) - f(x) \vert \le 1\) 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.
By assuming that $B(X)$ is a complete metric space, can you prove that $C _ B(X)$ is a complete metric space with respect to the $\infty$-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. \(\vert \vert f _ n - f \vert \vert _ \infty \le \varepsilon / 3\) Since $f _ n$ is continuous, there exists $\delta > 0$ s.t. \(\vert f _ n(x) - f _ n(a) \vert < \varepsilon/3\) for all $x \in B(a, \delta)$, so we have \(\begin{aligned} \vert f(x) - f(a) \vert &\le \vert f(x) - f _ n(x) \vert + \vert f _ n(x) - f _ n(a) \vert + \vert f _ n(a) - f(a) \vert \\\\ &< \varepsilon/3 + \varepsilon/3 + \varepsilon/3 \\\\ &= \varepsilon \end{aligned}\) so the limit is also continuous, hence $C _ B(X)$ is a closed subset of $C(X)$, so it is complete.
Proofs
Quickly prove that if a subspace of a complete metric space is complete if and only if is closed.
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.