Real Analysis: Global Metric Topology
Completeness
Motivation
Despite the fact that there are rational numbers whose squares are very close to $2,$ there is no rational number whose square is exactly $2.$ Thus, it appears that there is a kind of hole in the rational numbers where $\sqrt{2}$ should be. In contrast, the existence of $\sqrt{2}$ as a real number is not in doubt. The proof for the existence of $\sqrt{2}$ rests on the least upper bound property of $\mathbb{R},$ and it is this property than distinguishes the real numbers from the rational numbers by ensuring that no hole exists at not only $\sqrt{2},$ or even any other root, but anywhere else conceivable on the number line. This notion of "not having holes" is called completeness, and it can be generalized from the real numbers to metric spaces. However, to do so requires relying on something other than the least upper bound property, for metric spaces are not guaranteed to have orders imposed on them.
Completeness
To define completeness, we must first define the notion of a nested sequence of subsets. A sequence $\{S_n\}$ of subsets of a metric space $(X, d)$ is a nested sequence of subsets if it has the following properties:
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Each $S_n$ is closed, bounded, and nonempty.
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$S_{n+1} \subseteq S_n.$
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For every $\varepsilon > 0,$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(S_n) < \varepsilon$ for all $n \geq N.$
A metric space has the nested subset property if every nested sequence of subsets has an intersection with one value in it. In turn, a metric space is complete if it has the nested subset property. Accordingly, a subset $T$ of a complete metric space $(X, d)$ is complete if the subspace $(T, d)$ is complete.
The definition of completeness is given in terms of closed sets, rather than ordering properties, as closed sets can be constructed out of the elements of any metric space, while metric spaces are not guaranteed to have an order imposed on them to facilitate the least upper bound property. That the least upper bound property does not generalize even to $\mathbb{R}^n$ is reason enough to come up with an alternate definition!
Nested Interval and Nested K-Cell Theorems
Two special cases of the nested subset property are the nested interval theorem for $\mathbb{R}$ and the nested k-cell theorem for $\mathbb{R}^n.$ They follow immediately from the general cases, although their more geometric nature makes their direct proofs somewhat easier to visualize. Notably, they are both sufficient to prove the Bolzano-Weierstrass theorem. They are stated here in their entireties for (wait for it) completeness.
A nested sequence of intervals is a sequence of nonempty closed intervals $\{I_n\}$ of $\mathbb{R}$ with the following two properties:
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$I_{n+1} \subseteq I_n$
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For every $\varepsilon > 0,$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(I_n) < \varepsilon$ for all $n \geq N.$
The nested interval theorem states that if $\{I_n\}$ is a nested sequence of intervals, then their intersection $\bigcap\limits_{n=0}^{\infty} I_n$ is not empty and contains a single value.
Analogously, a nested sequence of k-cells is a sequence of nonempty closed k-cells $\{S_n\}$ in $\mathbb{R}^k$ with the following two properties:
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$S_{n+1} \subseteq S_n$
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For every $\varepsilon > 0,$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(S_n) < \varepsilon$ for all $n \geq N.$
The nested k-cell theorem states that if $\{S_n\}$ is a nested sequence of k-cells under the Euclidean metric, then their intersection $\bigcap\limits_{n=0}^{\infty} S_n$ is not empty and contains a single value.
Completeness of $\mathbb{R}$ and $\mathbb{R}^n$
Both $\mathbb{R}$ and $\mathbb{R}^n$ are complete. As a test of the quality of the generalized definition for metric spaces, though, we should check two things. First, we should check that our proof of the nested subset property makes use of the least upper bound property. And second, we should be able to swap the nested subset property for the least upper bound property on $\mathbb{R}$ and show that the least upper bound property is implied.
Equivalence to Cauchy Completeness
A metric space $(X, d)$ is also complete if and only if it fulfills the Cauchy Criterion. That is, $(X, d)$ is complete if every Cauchy sequence in $X$ converges to a point in $X.$ This alternate definition is more commonly used by textbooks because sequences of points are easier to visualize than sequences of somewhat abstract closed sets.
To get an idea of the equivalence between the least upper bound property of and the fulfillment of the Cauchy Criterion of $\mathbb{R},$ consider a sequence of rational numbers where the $n$th element of the sequence has the first $n$ digits of $\sqrt{2}.$ The set of elements in the sequence has a least upper bound, $\sqrt{2},$ which is also the limit of the sequence.
That $\mathbb{R}$ and $\mathbb{R}^n$ fulfill the Cauchy Criterion is proven in the later section on Cauchy sequences.
Problems
Show that the real numbers have the nested subset property.
Let $d$ be the Euclidean metric on $\mathbb{R}.$ Consider a nested sequence of real subsets $\{S_n\}.$ For $\varepsilon > 0,$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(S_m) < \varepsilon$ for every $m > N.$ Since each $S_m$ is closed, it follows that $\inf(S_m) \in S_m$ and $\sup(S_m) \in S_m.$ Set $A = \{a_m : a_m = \inf(S_m)\}$ and $B = \{b_m : b_m = \sup(S_m)\}.$ Since each $b_m$ is an upper bound for $A,$ by the least upper bound property, $\sup(A)$ exists. Since the least upper bound property implies the greatest lower bound property, the fact that each $a_m$ is a lower bound for $B$ shows that $\inf(B)$ exists.
We show that $\sup(A) = \inf(B).$ Proof by contradiction. Assume $\sup(A) \neq \inf(B).$ Then $d(\sup(A), \inf(B)) = c > 0.$ Pick $\delta = \frac{c}{2}.$ Then there exists some $M \in \mathbb{N}$ such that $\text{Diam}(S_m) < \delta$ for all $m > M.$ But this implies that $d(\inf(S_m), \sup(S_m)) \leq d(\sup(A), \inf(B)) < \frac{c}{2},$ which is a contradiction. Therefore $\sup(A) = \inf(B)$ after all.
Finally, we show that $y \in S_n$ for all $n \in \mathbb{N}.$ Proof by contradiction. Assume there exists some $N \in \mathbb{N}$ such that $y \notin S_N.$ Since $S_N$ is closed, it follows that $S_N^c$ is open. Therefore there exists some $r > 0$ such that $B_r(y) \in S_N^c.$ But then $y - \frac{r}{2}$ is an upper bound for $A,$ which is a contradiction, since $y = \sup(A).$ Therefore $y \in S_N$ after all. Since this is true of all $S_N,$ it follows that $y \in \bigcap\limits_{i=0}^{\infty} S_M \neq \varnothing.$
Equivalent definition: Show that if the least upper bound property on $\mathbb{R}$ is swapped for the nested subset property, then the least upper bound property is implied by the nested subset property.
Define $\mathbb{R}^*$ to have all the properties of $\mathbb{R}$ except that the least upper bound property has been replaced by the nested subset property.
Part 1: We construct a nested sequence of intervals.
Consider a set $A \subset \mathbb{R}^*$ with the property that $A$ has an upper bound, call it $b_0.$ Let $a_0$ not be an upper bound of $A.$ Then $a_0 \leq b_0.$ Construct the interval $I_0 = [a_0, b_0].$ We see that $I_0 \cap A$ is nonempty, as it contains an element greater than $a_0$ and less than or equal to $b_0.$ Let $c_0 = \frac{a_0 + b_0}{2}.$ If $c_0$ is an upper bound for $A,$ set $I_1 = [a_0, c_0],$ otherwise set $I_1 = [c_0, b_0].$ Repeat this process of bisection to form intervals $I_2, I_3,$ and so on.
Set $\delta_0 = \text{Diam}(I_0) = b_0 - a_0.$ Set $\delta_1 = \text{Diam}(I_1).$ Since $c_0$ is halfway between $a_0$ and $b_0,$ it follows that $\delta_1 = \frac{1}{2}\delta_0.$ In general, $\delta_{n+1} = \frac{1}{2}\delta_{n},$ and so $\delta_n = 2^{-n} \delta_0.$ Pick $\varepsilon > 0.$ By the Archimedean property on $\mathbb{R}^*,$ there exists some $n$ such that $2^{-n}\delta_0 < \varepsilon.$ It follows that $\{I_n\}$ forms a nested sequence of intervals. By completeness on $\mathbb{R}^*$ it follows that $\{I_n\}$ has a nonempty intersection and contains a single value, call it $x.$
Part 2: We show $x$ is an upper bound for $A.$
Proof by contradiction. Assume there is some $y \geq x$ such that $y$ is not an upper bound for $A.$ Set $\varepsilon = y - x.$ By the Archimedean property on $\mathbb{R}^*,$ there exists some $N \in \mathbb{N}$ such that $2^{-N}\delta_0 < \varepsilon.$ It follows that $|b_N - x| < \varepsilon,$ and therefore that $x < b_N < y.$ But this is a contradiction, since $b_N$ is an upper bound for $A.$ Therefore $x$ must have been an upper bound for $A$ after all.
Part 3: We show $x = \sup(A).$
Proof by contradiction. Assume there is some $y < x$ such that $y$ is an upper bound for $A.$ Set $\varepsilon = x - y.$ By the Archimedean property on $\mathbb{R}^*,$ there exists some $N \in \mathbb{N}$ such that $2^{-N}\delta_0 < \varepsilon.$ It follows that $|x - a_N| < \varepsilon,$ and therefore that $y < a_N < x.$ But this is a contradiction, since $a_N$ is not an upper bound for $A.$ It follows that $x$ must have been the least upper bound for $A$ after all.
Nested interval theorem: Show that if $\{I_n\}$ is a sequence of nested intervals, then $\bigcap\limits_{n=0}^{\infty} I_n$ is nonempty.
Consider the set $A = \{ a_n : n \in \mathbb{N}\}$ of lower bounds of $I_n.$ Note that each $b_n$ is an upper bound for $A.$ By the least upper bound property, it follows that $A$ has a least upper bound, call it $\sup(A) = x.$ Consider the interval $I_n = [a_n, b_n]$ for some $n \in \mathbb{N}.$ Since $x$ is an upper bound for $A,$ it follows that $a_n \leq x.$ Likewise, because each $b_n$ is an upper bound for $A$ and $x$ is the least such upper bound, it follows that $x \leq b_n.$ Therefore $x \in I_n.$ Since this is true of for any $I_n,$ it follows that $x \in \bigcap\limits_{n=0}^{\infty} I_n.$
Nested k-cell theorem: Let $\{S_n\}$ be a nested sequence of closed k-cells. Show that $\bigcap\limits_{n=0}^{\infty} S_n$ is nonempty.
Hint: Use the nested interval theorem.
Let $\{S_i\}$ be a sequence of nested k-cells where $S_i = \overline{K}_{n_i}(\mathbf{a}_i, \mathbf{b}_i)$ for $i \in \mathbb{N}.$ Define the sets $I_{i,j} = [a_{i,j}, b_{i,j}]$ for $1 \leq j \leq n.$ It follows that the sequence $\{I_{i,j}\}$ for each fixed $j$ is a sequence of nested intervals. By the nested interval theorem, $\bigcap\limits_{j=0}^{\infty} I_{i,j}$ contains a value $x_i.$ Therefore $\mathbf{x} = [x_1, \ldots, x_n] \in \bigcap\limits_{i=0}^{\infty} S_n.$
Show that if $T$ is a closed subset of a complete metric space $(X, d),$ then $T$ is complete.
Let $T$ be a closed subset of a complete metric space $(X, d),$ and let $\{S_n\}$ be a sequence of nested subsets of the metric subspace $(T, d).$ Since $T$ is closed in $X,$ it follows that each $S_n$ that is closed in $(T, d)$ is also closed in $(X, d).$ Since $X$ is complete, it follows that $\bigcap\limits_{n=0}^{\infty} S_n$ has a nonempty intersection. Therefore $T$ is complete.
Show that if $(Y, d)$ is a complete subspace of $(X, d),$ then $Y$ is closed in $X.$
Let $(Y, d)$ be a complete metric space, and let $(X, d)$ be a metric superspace of $(Y, d).$ If $\partial Y$ is empty, then $Y$ is closed in $X$ and the proof is complete. Otherwise, consider $x \in \partial Y,$ and consider the sequence of closed balls $\{\overline{B}_{1/n}(x)\}.$ We can see that intersection $\bigcap\limits_{n=0}^{\infty} \overline{B}_{1/n}(x) = x.$ Since each closed ball is closed in $X,$ it follows that each $\overline{B}_{1/n}(x) \cap Y$ is closed in $Y.$ It follows by the completeness of $Y$ that $\bigcap\limits_{n=0}^{\infty} \left(\overline{B}_{1/n}(x) \cap Y\right)$ is nonempty and therefore contains $x.$ Therefore $x \in Y,$ and so $\partial Y \subseteq Y$ and $Y$ is closed in $X.$
Show that if $(Y, d)$ and $(Z, d)$ are complete subspaces of $(X, d),$ then $(Y \cap Z, d)$ is also complete.
Since $Y$ and $Z$ are both complete, they are both closed in $X$. Because the intersection of closed sets is closed, it follows that $Y \cap Z$ is closed in $X.$ Thus $Y \cap Z$ is closed in $Y$. Since the closed subset of a complete metric space is complete, it follows that $Y \cap Z$ is complete.