Completeness I

The proof goes by noticing that the property of Cauchyness does not require a limiting value, and only depends on the elements in the sequence. So a sequence $(x_n)$ that has each element in $X$ is Cauchy in $Y$ iff it is Cauchy in $X$. The connection to closedness is that to be closed means to contain all of the limit points. So we start by considering $X$ as a subset of $Y$, and looking at its limit points and thus sequences which converge in $Y$ but whose elements are contained in $X$. Then since Cauchyness is equivalent to convergence in $Y$, we pass to Cauchy sequences, which then lets us consider sequences in $X$. The proof is short, and feels a little magical to me.

$\begin{aligned} \operatorname{cl}X \subseteq X &\iff \forall\, (x_n)\,\text{convergent in }Y\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_Y x_n \in X \\&\iff \forall\, (x_n)\,\text{Cauchy in }Y\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_Y x_n \in X \\&\iff \forall\, (x_n)\,\text{Cauchy in }X\text{ with } x_n\in X\,\,\forall n,\,\text{then} \, \lim_X x_n \text{ exists} \\&\iff X \text{ is complete.} \end{aligned}$

(Sketch). To show $(1)\Rightarrow(2)$, we simply take a sequence $x_n \in F_n$ and show it’s Cauchy, using the contracting condition. Completeness gives us a limit point, and since each set is closed, the limit point is in each $F_n$. A final application of the contracting condition gives us uniqueness.

On the other hand, given a Cauchy sequence $(x_n)$, we can define a contracting$(x_n)$ being Cauchy gives us the contracting condition.

sequence of closed sets by $F_n = \operatorname{cl}\{x_k: k\geq n\}$; then (2) gives us convergence.

(Sketch). To show $(1)\Rightarrow(2)$, we just note that the partial sums $s_n = \sum_{j=1}^n v_j$ form a Cauchy sequenceusing the absolute convergence of the norms

, and hence converge to some $v\in V$.

For $(2)\Rightarrow(1)$, consider a Cauchy sequence $(x_n)$ in $V$. By our earlier remarks, we only need to find a convergent subsequence. Choose $n_k$⊕ For a general Cauchy sequence, we don’t know how fast the terms get close together. So we exploit that they do get close to choose a subsequence where the terms get close exponentially fast.

such that if $m,n\geq n_k$ then $\|x_m - x_n\| \lt 2^{-k}$; wlog, we can take $n_k \gt n_{k-1}$ so that we have a legitimate subsequence. Note in particular, $\|x_{n_{k+1}} - x_{n_k}\| \lt 2^{-k}$. Now, we’ll write the telescoping sum $x_{n_k} - x_{n_1} = \sum_{k=1}^{n-1} x_{n_k+1} - x_{n_k}$ in order to use (2): since

$\begin{aligned} \sum_{k=1}^{\infty} \|x_{n_k+1} - x_{n_k}\| \leq \sum_{k=1}^\infty 2^{-k} = 1 \lt \infty,\end{aligned}$

2. tells us that $x_{n_k} - x_{n_1}=\sum_{k=1}^{n-1} x_{n_k+1} - x_{n_k} \xrightarrow{V} x\in V$.

Thus, $x_{n_k} \to x - x_{n_1} \in V$, and we’ve found a convergent subsequence to $(x_n)$.

(Proof of remark). We use the triangle inequality on the partial sums $s_n = \sum_{j=1}^n v_j$ to obtain

$\begin{aligned} \|s_n\| \leq \sum_{j=1}^n \|v_j\|.\end{aligned}$

and note that LHS and RHS here are convergent sequences of real numbersReverse triangle tells us $| \|s_n\| - \|v\| | \leq \|s_n - v\| \to 0$, so $\|s_n\| \xrightarrow{\mathbb{R}} \|v\|$.

so we can take the limit $n\to \infty$ to get the bound on $\|v\|$.

(Sketch). If $f$ had two fixed points $f(x) = x$ and $f(y) = y$, the contraction definition would give us a contradiction unless $x=y$.

To find the fixed point, we choose $x_0\in X$ arbitrarily, then make a sequence $x_n = f(x_{n-1})$. That is, $x_n = \underbrace{f\circ f \circ \dotsm \circ f}_{n\text{ times}} (x_0)$. ⊕ Since $f$ is contracting, if we apply it enough times it stops moving.

Since two consecutive terms are exponentially close by contraction:

$\begin{aligned} d(x_{n+1},x_n) = d(f(x_n), f(x_{n-1}))\leq \alpha d(x_n, x_{n-1})\leq \alpha^n d(x_1,x_0),\end{aligned}$

we can triangle inequality $d(x_n,x_m)$ by putting in all the consecutive terms, and estimate with a geometric series $\alpha^m\sum_{k=1}^\infty \alpha^k = \frac{\alpha^m}{1-\alpha}$ to find Cauchyness. Then completeness gives us convergence, and moreover,

$\begin{aligned} \lim x_{n} = \lim x_{n+1} = \lim f(x_n) = f( \lim x_n)\end{aligned}$

since contractions are continuous.

(Sketch). One only needs to apply uniform continuity to the definition of a Cauchy sequence. Uniformity is needed to compare the image of any two elements of the sequence far enough along.

(Sketch). We use the uniformity of the continuity twice here: once to map Cauchy to Cauchy, and once to get uniformity of our extension out of uniformity of $f$. If we only wanted a continuous extension, we wouldn’t need uniformity in the second usage, but we still need uniformity to map Cauchy to Cauchy. So we can’t use this method to prove that we can make continuous extensions from continuous functions. In fact, we cannot always extend continuous functionsIf we replace dense with closed and the codomain is $[-1,1]$, then we can extend continuous functions (this is the Tietze Extension Theorem), but this has nothing to do with completeness, so we won’t discuss this further here.

. For example, the set $A = (0,1)\subset \mathbb{R}$ is dense in $[0,1]$, and $f: A \to [-1,1],\, x \mapsto \sin(1/x)$ is continuous, but has no continuous extension (can’t define at zero).

To define $\tilde{f}$ on $x$, a limit point of $A$, we choose a sequence $(x_n)$ in $A$ which converges to $x$. This sequence is Cauchy, so $f(x_n)$ converges in $Y$. So we define our extension as $\tilde{f}(x) = \lim_Y f(x_n)$.

To show this is well-defined, given two sequences $(x_n)$ and $(y_n)$ in $A$ converging to $x$, we intertwine them to create $(z_n)$ in $A$ converging to $x$; then $\lim f(z_n)$ exists so its two subsequence $f(x_n)$ and $f(y_n)$ must share the same limit.

To show $\tilde{f}$ is uniformly continuous, we consider two limit points $x$ and $y$ with sequences $(x_n)$ and $(y_n)$ in $A$ converging to $x$ and $y$. We want to show if $x$ is close enough to $y$, then $\tilde{f}(x)$ is close to $\tilde{f}(y)$. For large $n$, $x_n$ is close to $y_n$, so by uniform continuity of $f$, $d_Y(f(x_n),f(y_n))$ is small. But by continuity of the metric,

$d(f(x_n),f(y_n)) \xrightarrow{n\to\infty} d(\tilde{f}(x), \tilde{f}(y)).$

Uniqueness follows from the uniqueness of limits: any two continuous extensions $h$ and $g$ must agree on $A$, and by continuity must then agree on limit points.

(Sketch). We need to show for any open set $O$, $O\cap \bigcap_{n=1}^\infty O_n$ is nonempty. We will inductively create a Cauchy sequence $(x_n)$ which will converge to a point in this intersection, using the density of $O_j$ at each step.

⊕ We are exploiting density and openness crucially here: near each point we have something in $O_n$, and near that something we have enough points to start afresh, and look for a point in $O_{n+1}$.

First, since $O\cap O_1$ is nonempty since $O_1$ is dense, and open since both sets are, we find some ball $B(x_1,\epsilon) \subset O\cap O_1$. Then we shrink the radius to some $r_1 \lt \epsilon$ to obtain $\operatorname{cl}B(x_1,r_1) \subset O\cap O_1$. This ball $B(x_1,r_1)$ serves as our "$O$’’ in the next step; we intersect with $O_2$, find some element and a ball around it in the intersection, and repeat. We can take the radius to shrink by at least $r_n \lt 1/n$ in each step, and recover a Cauchy sequence $(x_n)$ with

$x_n \in \operatorname{cl}(B(x_n,r_n)) \subset O_n \cap B(x_{n-1},r_{n-1}).$

⊕ This tail argument is why we bother messing with the closed balls earlier; we want to guarantee our limit is in the intersection.

Our shrinking radii $r_n \lt \frac{1}{n}$ yields that $(x_n)$ is Cauchy, and completeness gives a limit $x = \lim x_n$. Since the tail $\{ x_n: n\geq k\} \subset \operatorname{cl}(B(x_k,r_k))$ for each $k$, we have that the limit is in each closed ball too; each $\operatorname{cl}B(x_k,r_k) \subset O_k$, and the first ball is in $O$ so the limit $x\in O\cap \bigcap_{n=1}^\infty O_n$.

(Sketch). A set $A$ is dense iff $A^c$ has empty interior. Why?

• If $A$ is dense, then for $x\in A^c$, there exists $A\ni x_n \to x$. But then for any ball $B(x,\epsilon)$, there is some $x_n \in B(x,\epsilon)$, so $B(x,\epsilon) \not \subseteq A^c$.
• On the other hand, if $A$ has empty interior, then for any $x\in A$, $B(x,1/n) \not \subseteq A$, so we can take $x_n \in B(x,1/n) \cap A^c$. Then $x_n\to x$. For any $x\in A^c$, the constant sequence $x\to x$. So for any point $x\in X = A\cup A^c$, we can find a sequence in $A^c$ converging to $x$.

Using this, by taking complements of the $F_n$, applying BCT, then taking a complement again we obtain the result.

(Proof by BCT). If $\mathbb{Q}= \bigcap_{j=1}^\infty U_j$ for sets $U_j$ open in $\mathbb{R}$, then we could write

$\begin{aligned} \mathbb{R}= \mathbb{Q}\cup \mathbb{Q}^c = \bigcup_{n=1}^\infty \{r_n\} \cup \bigcup_{j=1}^\infty U_j^c.\end{aligned}$

Each of these sets is closed, and we have a countable union. So our last remark to BCT says they cannot all have empty interior. But the $\{r_n\}$ have empty interior, so some $U_j^c$ must have an interior; since $\mathbb{Q}$ is dense, we have $\mathbb{Q}\cap U_j^c\neq \emptyset$. But $U_j^c \subseteq \mathbb{Q}^c$, so $\mathbb{Q}\cap U_j^c = \emptyset$.