3.5Compact Subspaces of the Real Line

Let $X$ be a totally ordered set with the least upper bound property. In its order topology, closed intervals $[a, b] \subseteq X$ are compact subsets of $X$.

This proof is quite long, so we break it down intuitively. Consider any closed interval $[a, b]$ from the space $X$ in its order topology with the least upper bound property. Let $\mathcal{A}$ be an open cover for $[a, b]$ comprising subsets of $[a, b]$ that are open in its subspace topology (which coincides with its order topology). First we establish that, for any point $x \in [a, b)$, there is some element of this interval next to $x$ such that the area between them ($[x, y]$) can be covered by finitely many members of $\mathcal{A}$. Obviously we exclude $b$, since nothing comes after $b$. Knowing this, we know that there is some element after $a$ such that the area between them can be covered by finitely many members of $\mathcal{A}$. We let $C$ be all the elements of $[a, b]$ that satisfy this, and since it is nonempty, the value $c = \sup(C)$ is well-defined. For any member of $\mathcal{C}$ that contains $c$, since it is open, there is some nontrivial intersection with $C$, say at $z \in C$. By definition of $C$, we know $[a, z]$ can be finitely covered, and since $[z, c] \subseteq A$, the entire interval $[a, c] = [a, z] \cup [z, c]$ can be finitely covered, and so $c \in C$. Finally, to show $c = b$, we note that if $c < b$, then there is some $y \in (c, b]$ such that $[c, y]$ is finitely covered, and so $[a, y] = [a, c] \cup [c, y]$ is finitely covered, contradicting that $c$ is an upper bound for $C$. Thus $c = b$, completing the proof.

Every closed interval in $\bR$ is compact.

A subspace $A \subseteq \bR^n$ is compact if and only if it is closed and bounded under either the square or Euclidean metric.

Suppose $f: X \to Y$ is a continuous map from a compact set $X$ to an ordered set $Y$ in its order topology. Then there are $c, d \in X$ such that $f(c) \le f(x) \le f(d)$ for every $x \in X$.

Suppose $(X, d)$ is a metric space, and let $A \subseteq X$ is nonempty. Then, for any $x \in X$, we define

For a metric space $(X, d)$ and a fixed, nonempty subset $A \subseteq X$, the function $d(-, A)$ is continuous.

Let $(X, d)$ be a compact metric space together with any arbitrary open cover $\mathcal{A}$. Then there exists some $\delta > 0$, called a Lebesgue number, such that every nonempty subset of $X$ with diameter less than $\delta$ is contained in some member of the cover.

Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces. A map $f: X \to Y$ is said to be uniformly continuous if, for every $\epsilon > 0$, there exists some $\delta > 0$ such that for every $x_1, x_2 \in X$, it follows that

Suppose $f: X \to Y$ is a continuous map from the compact metric space $(X, d_X)$ to a target metric space $(Y, d_Y)$. Then $f$ is uniformly continuous.

Let $X$ be a topological space. A point $x \in X$ is said to be isolated if $\{x\}$ is open. Equivalently, $x$ is isolated if it is not a limit point of $X$, or if there exists an open neighborhood $O \subseteq X$ of $x$ that has no other points.

Suppose $X$ is a nonempty, compact, Hausdorff space. If $X$ has no isolated points, then it is uncountable.

Every nonempty compact subset of $\bR$ is uncountable.

1. Let $X$ be a totally ordered set with the least upper bound property. In its order topology, closed intervals of $X$ are compact subspaces.

   Intuition.

   Let $[a, b] \subseteq X$ be a closed interval, and let $\mathcal{A}$ be an open cover for this interval, where it takes open subsets of $[a, b]$ in its subspace topology. First, we show that for every $x \in [a, b)$, there is some $y > x$ such that $[x, y] \subseteq [a, b]$ can be covered by finitely many members of the cover $\mathcal{A}$ (in fact, by up to two!).

   Then we consider starting at $a$. Let $C$ be the set of all points $y > a$ such that $[a, y] \subseteq [a, b]$ can be covered by finitely many members of $\mathcal{A}$. We are guaranteed that $C \ne \emptyset$, since letting $x = a$ and applying the previous logic proves there is some element in $C$. Since it is non-empty and bounded, there is some $c = \sup(C)$. Let $A \in \mathcal{A}$ be a member of the cover such that $c \in A$. By $A$ being open, it follows there is some $d < c$ such that $c \in (d, c] \subseteq A$. Note that there is some $z \in C$ such that $d < z < c$, otherwise $d$ would be a smaller upper bound than $C$. Then $[a, c] = [a, z] \cup [z, c]$, where the first part is covered by finitely many members of $\mathcal{A}$ as per the definition of $C$, and the second part is covered by finitely many members of $\mathcal{A}$ since $[z, c] \subseteq A$—thus $c \in C$.

   Now we show $c = b$. For the sake of contradiction, suppose $c < b$. Following the logic of the previous paragraph, there is some $A \in \mathcal{A}$ such that $c \in A$, and since $A$ is open, there is some $x \in (c, b]$ such that $[c, x] \subseteq A$. Then $[a, x] = [a, c] \cup [c, x]$ can be finitely covered, meaning $x \in C$, contradicting that $c$ is an upper bound. Thus $c = b$.

2. As a consequence, taking $X = \bR$ means are closed intervals in $\bR$ are compact.

3. Compact iff. Closed and Bounded in $\bR^n$. A subspace $A \subseteq \bR^n$ is compact if and only if it is closed and bounded under the Euclidean or square metrics.

   Intuition.

   Note that a subspace $A \subseteq \bR^n$ is bounded under the Euclidean metric if and only if it is bounded under the square metric, since

   $$\rho(x, y) \le d(x, y) \le \sqrt{n} \rho(x, y)$$

   for every $x, y \in \bR^n$.

   $(\Longrightarrow)$: Suppose $A$ is compact. Then $A$ is a compact subspace of the Hausdorff space $\bR^n$, and so $A$ is closed. Let $\{B_{\rho}(\vb{0}_n, m)\}_{m \in \bZ_+}$ be an open cover for $\bR^n$, then choose a finite subcover that covers $A$. Then the union of all members of the finite subcover is contained in a box, and since $A$ is inside this box, $A$ is bounded as well.

   $(\Longleftarrow)$: Suppose $A$ is closed and bounded under $\rho$. Suppose that $N \in \bR$ is such that $\rho(x, y) \le N$ for each $x, y \in A$. Fix some $x_0 \in A$, then let $\rho(x_0) = b$. Then note $\rho(x, \vb{0}_n) \le \rho(x, x_0) + \rho(x_0, \vb{0}_n) = N + b$ for every $x \in A$, and so $A \subseteq B_{\rho}(\vb{0}_n, N + b)$. Since $A$ is a closed subset of the compact box, $A$ is compact as well.

4. Extreme Value Theorem. Let $f: X \to Y$ be a continuous map from a topological space $X$ to a totally ordered set $Y$ in its order topology. If $X$ is compact, there exist $c, d \in X$ such that $f(c) \le f(x) \le f(d)$ for every $x \in X$.

   Intuition.

   This theorem essentially states that $f(X)$ has a minimum and maximum. We prove $f(X)$ contains its maximum, since the proof that it contains its minimum is analogous. For the sake of contradiction, we suppose $f(X)$ does not contain a maximum. Then note $\{(-\infty, a)\}_{a \in f(X)}$ is a set of open subsets of $Y$. Since $f(X)$ has no maximum, for each $y \in f(X)$, there is some $a > y \in f(X)$, so this collection forms an open cover for $f(X)$. Since $X$ is compact, $f(X)$ is also compact by continuity of $f$. Thus there is a finite subcover

   $$\{(-\infty, a_1), \dots, (-\infty, a_n)\}$$

   that covers $f(X)$. Note $a = \max\{a_1, \dots, a_n\}$ exists and should be an element of $A$, but it is not contained in the finite subcover, a contradiction.

5. For a metric space $(X, d)$ and some nonempty subset $A \subseteq X$, we define the distance of a point $x \in X$ from $A$ by $d(x, A) = \inf_{a \in A}\{d(x, a)\}$. Colloquially, this is the shortest distance from the point to the set $A$. Compare this to the diameter of $A$, which is colloquially the largest distance between any two points in $A$.

6. The function $d(-, A)$ is continuous.

   Intuition.

   Let $x, y \in X$ be arbitrary points. Note then that

   $$d(x, A) \le d(x, a) \le d(x, y) + d(y, a)$$

   for every $a \in A$. Notably, we seek to minimize $d(y, a)$, such that

   $$d(x, A) \le d(x, y) - d(y, a) \le d(x, y) - \inf_{a \in A} \{d(y, a)\} \implies d(x, A) - d(y, A) \le d(x, y).$$

   Interchanging the role of $x, y$ yields $|d(x, A) - d(y, A)| \le d(x, y)$, implying $d(-, A)$ is Lipschitz and thus continuous.

7. Lebesgue Covering Lemma. Suppose $(X, d)$ is a compact metric space together with an open cover $\mathcal{A}$. Then there exists some $\delta > 0$, called a Lebesgue number, such that every set whose diameter is less than $\delta$ is contained in one member of the cover.

   Intuition.

   Since $X$ is compact, it admits a finite subcover $\{A_k\}_{k = 1}^n$, where each $A_k \in \mathcal{A}$. Let $C_k = X \setminus A_k$ for each $k$, then define the function $f: X \to \bR$ by

   $$f(x) = \frac{1}{n} \sum_{k = 1}^n d(x, C_k).$$

   We demonstrate that $f(x) > 0$ for every $x \in X$. For a fixed $x \in X$, there is some $k$ such that $x \in A_k$. Since $A_k$ is open, there is some $\epsilon > 0$ such that $x \in B(x, \epsilon) \subseteq A_k$. Thus $d(x, C_k) \ge \epsilon$ and so $f(x) \ge \epsilon/n$. Since $X$ is compact, $f$ attains a minimum value $\delta > 0$ (where $\delta > 0$ since $f(x) > 0$) by EVT. We show this is our desired Lebesgue number.

   Consider any set $B$ whose diameter is $\delta$. Then there is some $x_0 \in B$ such that $B \subseteq B(x_0, \delta)$. Then note

   $$\delta \le f(x_0) \le d(x, C_m),$$

   where $m$ is chosen to maximize the value of $d(x, C_k)$ for each $k$. Then $B(x_0, \delta) \subseteq A_m$, where $A_m = X \setminus C_m$, implying $B \subseteq A_m \in \mathcal{A}$, completing the proof.

8. Uniform Continuity Theorem. Suppose $f: X \to Y$ is a continuous map from a compact metric space $(X, d_X)$ to a target metric space $(Y, d_Y)$. Then $f$ is uniformly continuous, meaning that for any $\epsilon > 0$, there is some $\delta > 0$ such that

   $$d_X(x_1, x_2) < \delta \implies d_Y(f(x_1), f(x_2)) < \epsilon$$

   for every $x_1, x_2 \in X$.

   Intuition.

   Fix $\epsilon > 0$. Let $\mathcal{A} = \{B_{d_Y}(y, \epsilon/2)\}_{y \in Y}$ be an open cover of $f(X)$ from open subsets of $Y$. Then note $f^{-1}(\mathcal{A})$ forms an open cover of $X$. Since $X$ forms a compact metric space, there is a Lebesgue number $\delta$. Let $x_1, x_2 \in X$ be such that $d_X(x_1, x_2) < \delta$. Then $\{x_1, x_2\}$ is a set whose diameter is less than $\delta$, meaning $\{x_1, x_2\} \subseteq A$ for some $A \in \mathcal{A}$. This implies $\{x_1, x_2\} \subseteq f^{-1}(B_{d_Y}(y, \epsilon/2))$ for some $y \in Y$, and so $\{f(x_1), f(x_2)\} \subseteq B_{d_Y}(y, \epsilon/2)$. Thus $d_Y(f(x_1), f(x_2)) < \epsilon$, completing the proof.

9. For any topological space $X$, a point $x \in X$ is said to be isolated if $\{x\}$ is open. Equivalently, $x$ is isolated if an open neighborhood of $x$ exists that comprises only $x$, or if $x$ is not a limit point of $X$.

10. Suppose $X$ is a nonempty, compact, Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

    Intuition.

    The intuition behind this proof is to construct a sequence of nested, open sets $V_1 \supseteq V_2 \supseteq \cdots$ with the finite intersection property, such that $x_n \in V_n$ but $x_n \notin \bar{V}_{n + 1}$ for every $n \in \bZ_+$, where $x_n \in X$. Then, taking the closure of each $V_k$ (note that if $V_k \subseteq V_{k + 1}$, then $\bar{V}_k \subseteq \bar{V}_{k + 1}$), we yield a sequence of nested, closed sets with the finite intersection property. Thus $x \in \bigcup_{k \in \bZ_+} \bar{V}_k$, but $x \ne x_n$ for each $n$, completing the proof that $X$ is uncountable.

    Let $X$ be a nonempty, compact, Hausdorff space with no isolated points. Let $U \subseteq X$ be a nonempty, open set and let $x \in X$ be arbitrary. We seek to show there is a nonempty, open subset $V \subseteq U$ such that $x \notin \bar{V}$. Choose a point $y \in U$ such that $x \ne y$—if $x \in U$, then $U$ necessarily contains a distinct point $y$ to prevent $\{x\}$ from being open (since $X$ has no isolated points); otherwise if $x \notin U$, then $y$ is guaranteed since $U$ is nonempty. By Hausdorffness of $X$, there exists open neighborhoods $W_1, W_2 \subseteq X$ of $x$ and $y$ repsectively. Then $V = W_2 \cap U$ is a nonempty, open subset of $U$. Note $x \notin \bar{V}$ since $W_1$ is an open neighborhood of $X$ that doesn't intersect $W_2$, so $x$ is not a limit point of $V$.

    Consider the subset $\{x_n\}_{n \in \bZ_+} \subseteq X$. Let $V_1 = X$, and let $V_{k + 1} \subseteq V_k$ be nonempty and open such that $x \notin V_{k + 1}$ (existence is proved by the last paragraph). Then the closure of each $V_k$ forms a nested sequence of closed sets with the finite intersection property, so there is some $x \in \cup_{k \in \bZ_+} \bar{V}_k$. Note that, for each $n \in \bZ_+$, $x_n \notin \bar{V}_n$ and so $x \ne x_n$. Thus $x$ is an element of $X$ outside this subset, making $X$ uncountable.

11. As a direct corollary, it follows that $\bR$ is uncountable.

Suppose $X$ is a totally ordered set in its order topology. Show that if every closed interval of $X$ is compact, then $X$ has the least upper bound property.