Intersections of non-compact sets can be badly behaved
We showed in lecture that if $\set{K_\alpha}$ is a collection of compact subsets of $M$ so that the interesction of every finite subcollection is non-empty, then the intersection of all the $K_\alpha$ is non-empty.
- Show that this may fail if the $K_\alpha$ are merely assumed to be closed.
- Show that this may fail if the $K_\alpha$ are merely assumed to be bounded.
- Suppose that $\set{K_\alpha \mid \alpha\in\mathcal{I}}$ is a collection of compact subsets of $M$ so that the intersection $K_\alpha\cap K_\beta$ is non-empty for each $\alpha, \beta \in \mathcal{I}$. Show that it may nonetheless be the case that
\[\bigcap_\alpha K_\alpha = \emptyset.\]
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Compactness in the rational numbers
Let $S = \set{q \in \Q \mid q^2 \lt 2}$.
Show that $S$ is closed and bounded as a subset of the metric space $\Q$, but is not compact.
Also, determine if $S$ is open as a subset of $\Q$.
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Open sets in a product space
Recall that if $(S, d_S)$ and $(T, d_T)$ are metric spaces, then the functions
\begin{align*}
d_1(\mathbf{x}, \mathbf y) &= d_S(x_1, y_1) + d_T(x_2, y_2) \\
d_2(\mathbf{x}, \mathbf{y}) &= \sqrt{d_S(x_1, y_1)^2 + d_T(x_2{\color{red},}y_2)^2}, \text{ and}\\
d_\infty(\mathbf x, \mathbf y) &= \max(d_S(x_1, y_1), d_T(x_2, y_2))
\end{align*}
are all metrics on $S\times T$.
Let $r \gt 0$ and $\mathbf x \in \color{red}{S\times T}$.
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Prove that there is some $s \gt 0$ so that
\[B_s^{d_1}(\mathbf x) \subseteq B_r^{d_2}(\mathbf x).\]
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Prove that there is some $s \gt 0$ so that
\[B_s^{d_2}(\mathbf x) \subseteq B_r^{d_\infty}(\mathbf x).\]
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Prove that there is some $s \gt 0$ so that
\[B_s^{d_\infty}(\mathbf x) \subseteq B_r^{d_1}(\mathbf x).\]
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Prove that if $G \subseteq \color{red}{S\times T}$ is open with respect to any of the three metrics, then it is open with respect to all of them.
(Note that there is no reason to find the best $s$ for a given $r$, just any one that works. It may be interesting to think about what the best is, though. Consider also what this says about drawing diamonds inside circles inside squares.
Also note that this applies to $\R^2$, and so by induction, to $\R^n$.)
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Variations on coverings
A metric space $(M, d)$ is called totally bounded if for every $\epsilon \gt 0$ there is a finite list of points $x_1, \ldots, x_n \in M$ so that every $y \in M$ is within $\epsilon$ of at least one $x_i$.
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Prove that compact metric spaces are totally bounded.
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Prove that a totally bounded metric space need not be compact.
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Suppose that $(K, d)$ is a compact metric space, and $\epsilon \gt 0$.
Prove that there is a subset $S \subseteq K$ so that for every $y \in K$ there is $x \in S$ with $d(x, y) \lt \epsilon$, and moreover so that for any distinct $x, w \in S$ we have $d(x, w) \gt \frac\epsilon2$.
(Hint: start with a cover by finitely many $\frac\epsilon2$ balls.)
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
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Products of compact sets
Suppose that $(X_1, d_1), (X_2, d_2)$ are compact metric spaces.
Let $K = X_1 \times X_2$ be equipped with the metric $$d\paren{(x_1, x_2), (y_1, y_2)} = \sqrt{d_1(x_1, y_1)^2 + d_2(x_2, y_2)^2}.$$
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Suppose $G \subseteq K$ is open and $(x_1, x_2) \in G$.
Show that for some $r \gt 0$,
$$B_r^{X_1}(x_1) \times B_r^{X_2}(x_2) = \set{(y_1, y_2) \in K \middle\vert d_i(x_i, y_i) \lt r \text{ for } i = 1, 2} \subseteq G.$$
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Let $\mathscr{U}$ be an open cover of $K$, fix $x \in X_1$, and write
$$\mathscr{U}_x = \set{ G \subseteq X_2 \middle\vert \begin{aligned} G \text{ is open, and } \exists r \gt 0 \text{ and } \\ E \in \mathscr{U} \text{ so that } B_r^{X_1}(x) \times G \subseteq E \end{aligned} }.$$
Prove that $\mathscr{U}_x$ is an open cover of $X_2$.
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Prove that there is $r_x \gt 0$ and a finite set
$$\mathscr{V}_x \subseteq \mathscr{U}$$
which is an open cover of $B_{r_x}(x) \times X_2 \subseteq K$.
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Prove that $K$ is compact.
(Hint:
$\set{B_{r_x}(x) \mid x \in X_1}$
covers $X_1$.)
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Prove that if $(X_1, d_1), (X_2, d_2), \ldots, (X_n, d_n)$ are compact metric spaces, then so is $X_1 \times X_2 \times \cdots \times X_n$ when equipped with the metric
$$d(\mathbf{x}, \mathbf{y}) = \sqrt{\sum_{i=1}^n d_i(x_i, y_i)^2},$$
where $\mathbf{x} = (x_1, \ldots, x_n)$ and $\mathbf{y} = (y_1, \ldots, y_n)$.
More compact products
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Repeat the above, but using the metric $d_\infty(\mathbf{x}, \mathbf{y}) = \max\paren{d_i(x_i, y_i)}$ instead.
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Suppose that $\mathcal{I}$ is an infinite set, and that for each $i \in \mathcal{I}$, $(X_i, d_i)$ is a compact metric space with the property that for all $x, y \in X_i$, $d_i(x, y) \leq 1$.
Let
$$X = \prod_{i\in\mathcal{I}} X_i := \set{ f : \mathcal{I} \to \bigcup_{i\in\mathcal{I}} X_i \,\middle\vert\, \forall\,i \in \mathcal{I}, f(i) \in X_i }.$$
Then $X$ is a metric space when equipped with the metric
$$d_\infty(f, g) = \sup\set{d_i(f(i), g(i)) \mid i \in \mathcal{I}}.$$
Prove that $X$ is compact, or show by example that this need not be the case.
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Suppose that for each $i \in \N$, $(X_i, d_i)$ is a compact metric space.
Designate a fixed "base point" $0_i \in X_i$ for each $i$, and let
$$Y = \set{ f : \N \to \bigcup_{i\in\N} X_i \,\middle\vert\, \forall\,i\in\N, f(i) \in X_i \text{ and } \sum_{i=1}^\infty d_i(f(i), 0_i)^2 \lt \infty}.$$
Then $Y$ is a metric space when equipped with the metric
$$d(f, g) = \sqrt{\sum_{i=1}^\infty d_i(f(i), g(i))^2}.$$
Prove that $Y$ is compact, or show by example that this need not be the case.
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Intersections of non-compact sets can be badly behaved
Find an example that shows the statement in problem 1 may fail if the $K_\alpha$ are closed and bounded, but not necessarily compact. (Hint: for this, you will need to work in a metric space other than $\R^n$.)
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Open sets in higher-dimensional Euclidean spaces
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Define $d_1, d_2, d_\infty$ on $\R^n$ analogously to how they are defined on $\R^2$.
Prove that if $G \subseteq \R^n$ is open with respect to any of the three metrics, then it is open with respect to all of them.
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Let $p \gt 1$ and define the metric $d_p$ on $\R^n$ by $d_p(\mathbf x, \mathbf y) = \paren{\sum_{i=1}^n \abs{x_i-y_i}^p}^{1/p}$.
Prove that $G \subseteq \R^n$ is open with respect to $d_p$ if and only if it is open with respect to $d_1$.
(Note that there is no reason to find the best $s \gt 0$ for a given $r$, just any one that works. It may be interesting to think about what the best is, though. Consider what this says about drawing diamonds inside circles inside squares.)
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Unbounded sets and limit points
Let $(M, d)$ be a metric space, and suppose that $E \subseteq M$ is not bounded.
Prove that $E$ has an infinite subset with no limit points in $M$.
Use this to show that compact sets are bounded.
Note that this gives another way to prove that compact sets are bounded.
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Covers by open balls
Suppose that $K$ is a subset of a metric space $(M, d)$ with the property that whenever $\mathscr{U}$ is an open cover of $K$ with the property that every set in $\mathscr{U}$ is a ball, it follows that $\mathscr{U}$ has a finite subcover.
Is this enough to guarantee that $K$ is compact?
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Disjoint open sets
Recall that a set $S$ is called countable if there is an injective (i.e., one-to-one) function $S \hookrightarrow \N$.
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Prove that if $\mathscr{V}$ is a collection of disjoint open subsets of $\R^n$, then $\mathscr{V}$ is countable.
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Prove that if $G \subseteq \R$ is open, then $G$ is the countable union of disjoint open intervals (counting $(-\infty, b), (a, \infty)$, and $(-\infty, \infty)$ as valid intervals).
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An iterative question
Let $S \subseteq \R$, and consider the collection $\mathscr{S}$ of subsets of $\R$ which can be formed by applying some sequence of complements and closures to $S$.
For example, if $S = (0, 1)$, then $\mathscr{S} = \set{(0, 1), [0, 1], \R\setminus(0, 1), \R \setminus[0,1]}$; taking further complements or closures of these sets produces no new sets.
Determine whether or not $\mathscr{S}$ can be infinite.
If it must be finite, determine if it can be arbitrarily large.
Provide either a set $S$ so that $\mathscr{S}$ is infinite, a sequence of sets $S_n$ so that the corresponding $\mathscr{S}_n$ has size at least $n$, or a set $S$ so that $\mathscr{S}$ is as large as possible.
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More covers
Use Zorn's Lemma to show that 4C is true in any metric space, not merely compact ones.
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The reals are closed
Suppose $(M, d)$ is a metric space so that $\R \subseteq M$ and for any $x, y \in \R$, $d(x, y) = \abs{y-x}$.
Prove that $\R$ is closed as a subset of $M$.