Real Analysis: Global Metric Topology

Problems

1. Equivalent definition: Show that if the least upper bound property on $\mathbb{R}$ is swapped for the Cauchy criterion, then the least upper bound property is implied.

   Define $\mathbb{R}^*$ to have all the properties of $\mathbb{R}$ except that the least upper bound property has been exchanged for the Cauchy criterion. We show that $\mathbb{R}^*$ has the least upper bound property.

   Part 1: We construct a sequence $\{a_n\}$ whose elements are not upper bounds of $A$ and a sequence $\{b_n\}$ whose elements are upper bounds of $A.$

   Let $A$ be a nonempty subset of $\mathbb{R}^*$ that is bounded above by $b_1,$ and let $a_1 \in \mathbb{R}^*$ such that $a_1$ is not an upper bound of $A.$ Set $\delta_1 = b_1 - a_1,$ the distance between $a_1$ and $b_1.$ Consider $c_1 = \frac{a_1 + b_1}{2}.$ If $c_1$ is an upper bound for $A,$ then set $a_2 = a_1$ and $b_2 = c_1.$ Otherwise, set $a_2 = c_1$ and $b_2 = b_1.$ In either event, set $\delta_2 = b_2 - a_2 = \frac{1}{2}\delta_1.$ Repeat this process of halving to form the two sequences $\{a_n\}$ and $\{b_n\}.$ Note that $\delta_n = 2^{-n+1}\delta_1.$

   Part 2: We show that $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences.

   Note that each $b_n$ is an upper bound for $A$ and therefore also for $\{a_n\},$ and note that no $a_n$ is an upper bound for $A,$ and thus a low bound for $\{b_n\}.$

   Pick $\varepsilon > 0.$ By the Archimedean property on $\mathbb{R}^*,$ there exists an $N \in \mathbb{N}$ such that $2^{-N}\delta_1 < \varepsilon.$ It follows that for any $n, m > N$ where $n < m$ that $|a_n - a_m| < |a_n - b_n| = 2^{-N}\delta_1 < \varepsilon.$ Therefore $\{a_n\}$ is a Cauchy sequence. By the Cauchy criterion on $\mathbb{R}^*,$ it follows that $\{a_n\}$ converges to a point $L_1.$

   Likewise, pick $\varepsilon > 0.$ By the Archimedean property on $\mathbb{R}^*,$ there exists an $N \in \mathbb{N}$ such that $2^{-N}\delta_1 < \varepsilon.$ It follows that for any $n, m > N$ where $n < m$ that $|b_n - b_m| < |a_n - b_n| = 2^{-N}\delta_1 < \varepsilon.$ Therefore $\{b_n\}$ is a Cauchy sequence. By the Cauchy criterion on $\mathbb{R}^*,$ it follows that $\{b_n\}$ converges to a point $L_2.$

   Part 3: We show that $\lim\limits_{n \rightarrow \infty} a_n = L = \lim\limits_{n \rightarrow \infty} b_n.$

   Proof by contradiction. Set $\varepsilon = L_2 - L_1.$ Since $\{a_n\}$ and $\{b_n\}$ converge, there exists an $N_1 \in \mathbb{N}$ such that $|a_n - L_1| < \frac{\varepsilon}{4}$ and $|b_n - L_2| < \frac{\varepsilon}{4}$ whenever $n \geq N_1.$ Likewise, by the Archimedean property on $\mathbb{R}^*,$ there exists an $N_2 \in \mathbb{N}$ such that $2^{-N+1}\delta_1 < \frac{\varepsilon}{4}.$ Set $N = \max\{N_1, N_2\}.$ Then $|a_n - b_n| < \frac{\varepsilon}{4}$ and $|a_n - L_1| < \frac{\varepsilon}{4}$ and $|b_n - L_2| < \frac{\varepsilon}{4}$ whenever $n > N.$ By the triangle inequality, this means that $|L_2 - L_1| \leq |L_1 - a_n| + |a_n - b_n| + |b_n - L_2| = \frac{3\varepsilon}{4}.$ But this is a contradiction. Therefore the limits must have been identical after all.

   Set $L = L_1 = L_2.$

   Part 4: We show that $L$ is an upper bound of $A.$

   We use trichotomy to show that $L$ cannot be otherwise. Assume there is a $c \in A$ such that $L < c.$ Set $\varepsilon = c - L.$ Then there exists an $N \in \mathbb{N}$ such that $|b_n - L| < \varepsilon$ whenever $n \geq N.$ This implies that $L < b_N < c$ But this is a contradiction, since $b_N$ is an upper bound of $A.$ Therefore $L \geq c$ after all.

   Part 5: We show that $L = \sup(A).$

   Next we show that $L$ is the least upper bound of $A.$ Proof by contradiction. Assume there is some $x < L$ such that $x$ is an upper bound of $A.$ Let $\varepsilon = L - x.$ Then there exists an $N \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ whenever $n \geq N.$ It follows that $x < a_N \leq L.$ But this is a contradiction, since $a_N$ is not an upper bound of $A.$ Therefore $L$ is the least upper bound of $A$.

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2. 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 $y \in \bigcap\limits_{i=0}^{\infty} S_M \neq \varnothing.$

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3. 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.

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4. Equivalent definition, Part 1: Show that if a metric space is complete, then it has the nested subset property.

   Let $(X, d)$ be a complete metric space, and let $\{S_n\}$ be a nested sequence of subsets.

   Part 1: Construct Cauchy sequences out of points of each $S_n.$

   From each $S_n,$ select two points $a_n, b_n \in S_n.$ It follows that $d(a_n, b_n) \leq \text{Diam}(S_n).$ Since $S_{n+1} \subseteq S_n,$ it also follows that $\text{Diam}(S_{n+1}) \leq \text{Diam}(S_n),$ and so $d(a_n, a_m) \leq \text{Diam}(S_n)$ for all $m > n.$ Since $\{S_n\}$ is a nested sequence of subsets, for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(S_n) < \varepsilon$ for all $n > N.$ Thus $d(a_n, a_m) < \varepsilon$ and $d(b_n, b_m) < \varepsilon$ for $n, m \geq N.$ It follows that $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences. By completeness on $(X, d),$ $\{a_n\}$ and $\{b_n\}$ converge. Let $\lim\limits_{n \rightarrow \infty} a_n = L_a$ and $\lim\limits_{n \rightarrow \infty} b_n = L_b.$

   Part 2: Show that $L_a = L_b.$

   Proof by contradiction. Assume $L_a \neq L_b.$ Set $\varepsilon = d(L_a, L_b) > 0.$ Then there exists an $N_1 \in \mathbb{N}$ such that $d(a_n, b_n) < \frac{\varepsilon}{4}$ whenever $n \geq N_1.$ Likewise, there exists an $N_2 \in \mathbb{N}$ such that $d(a_n, L_a) < \frac{\varepsilon}{4}$ whenever $n > N_2,$ as well as an $N_3 \in \mathbb{N}$ such that $d(b_n, L_b) < \frac{\varepsilon}{4}$ whenever $n > N_3.$ Set $N = \max\{N_1, N_2, N_3\}.$ It follows by the triangle inequality that $d(L_a, L_b) \leq d(L_a, a_n) + d(a_n, b_n) + d(b_n, L_b) = \frac{3\varepsilon}{4}.$ But this is a contradiction. Therefore $L_a = L_b$ after all.

   Let $L = L_a = L_b.$ 

   Part 3: Show that $L \in \bigcap\limits_{n=0}^{\infty} S_n.$

   Proof by contradiction. Assume $L \notin \bigcap\limits_{n=0}^{\infty} S_n.$ Then there is some $N_1 \in \mathbb{N}$ such that $L \notin S_{N_1}.$ Therefore $L \in S_{N_1}^c.$ Since $S_{N_1}$ is closed, it follows that $S_{N_1}^c$ is open. Therefore there exists some $r > 0$ such that $B_r(L) \subseteq S_{N_1}^c.$ Pick $\varepsilon = r.$ Then there exists some $N_2 \in \mathbb{N}$ such that $d(a_n, L) < \varepsilon$ whenever $n > N_2.$ Set $N = \max\{N_1, N_2\}.$ When $n > N,$ we see that $d(a_n, L) < r,$ and so $a_n \in B_r(L).$ But this implies that $a_n \in S_{N_1}^c,$ a contradiction since $a_n \in S_n \subseteq S_{N_1}.$ Therefore $L \in \bigcap\limits_{n=0}^{\infty} S_n$ after all.

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5. Equivalent definition, Part 2: Show that if a metric space has the nested subset property, it is complete.

   Let $(X, d)$ be a metric space with the nested subset property. Let $\{a_n\}$ be a Cauchy sequence.

   Part 1: Construct a nested sequence of closed subsets.

   Since $\{a_n\}$ is Cauchy, it is bounded. It follows that for every $n \in \mathbb{N},$ there exists an $r_n \in \mathbb{R}$ such that $r_n = \text{Diam}(\{a_m : m \geq n\}).$ Since subsequent terms in a sequence form a subset of their union with the preceding term, it follows that $r_{n+1} \leq r_n.$ Therefore $\overline{B}_{r_m}(a_m) \subseteq \overline{B}_{r_n}(a_n)$ for all $m \geq n.$ Since $\{a_n\}$ is Cauchy, for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \varepsilon$ for all $n, m \geq N.$ Therefore $r_n \leq \varepsilon,$ and so $\{\overline{B}_{r_n}(a_n)\}$ forms a nested sequence of subsets. By the nested subset property on $(X, d),$ the intersection $\bigcap\limits_{n=0}^{\infty} \overline{B}_{r_n}(a_n)$ is nonempty and contains a single value, call it $L.$

   Part 2: Show that the $\lim\limits_{n \rightarrow \infty} a_n = L.$

   Since $\{a_n\}$ is Cauchy, for any $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \varepsilon$ whenever $n, m \geq N.$ By construction, $\text{Diam}(\overline{B}_{r_n}(a_n)) < \varepsilon$ as well. Since $L \in \overline{B}_{r_n}(a_n),$ it follows that $d(a_n, L) < \varepsilon.$ Therefore $\lim\limits_{n \rightarrow \infty} a_n = L.$

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6. 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.

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7. 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 $X.$

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8. 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.

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