2.9Metric Topology

Proof.

We verify this forms a basis for $X$.

1. Covering. For any $x \in X$, note $B_1(x)$ is in the basis, and so $X \subseteq \bigcup_{B \in \mathcal{B}} B$ as desired.

2. Local Refinement. Let $B_1, B_2$ be arbitrary elements from the basis, and let $x \in B_1 \cap B_2$. Thus there must be some $\delta_1, \delta_2$ such that $B_{\delta_1}(x) \subseteq B_1$ and $B_{\delta_2}(x) \subseteq B_2$. Then note $B_{\min\{\delta_1, \delta_2\}}(x) \subseteq B_1 \cap B_2$, which is also a basic element, completing the proof.

❦

Proof.

The proof that $\bar{d}$ is a metric is trivial. Let $\Tau_d, \Tau_{\bar{d}}$ be the topologies generated by $d$ and $\bar{d}$ respectively. It is obvious that $\Tau_{\bar{d}} \subseteq \Tau_d$, so we prove that $\Tau_d \subseteq \Tau_{\bar{d}}$. Let $B \in \Tau_d$. By definition of $\Tau_d$, for each $x \in B$, we can choose a basis ball about $x$ contained in $B$; if its radius is $1$ or greater, we can simply shrink it. So let $B_x$ be a ball about $x$ such that its radius is less than $1$ and $x \in B_x \subseteq B$. Since its radius is less than $1$, the ball $B_x$ is identical under $d$ and $\bar{d}$, meaning it is a basic open set of $\Tau_{\bar{d}}$. Then $B = \bigcup_{x \in B} B_x$, a union of basic open sets of $\Tau_{\bar{d}}$, completing the proof. ❦

Proof.

$(\Longrightarrow)$: Suppose $\Tau \subseteq \Tau'$. Then, for any $x \in X$ and $\epsilon > 0$, we have that $B_d(x, \epsilon) \in \Tau$. Since $\Tau \subseteq \Tau'$, $B_d(x, \epsilon) \in \Tau'$. Thus, for $x \in B_d(x, \epsilon)$, by definition of the basis, there is some $\delta > 0$ such that $x \in B_{d'}(x, \delta) \subseteq B_d(x, \epsilon)$.

$(\Longleftarrow)$: Let $O \in \Tau$—we seek to show $O \in \Tau'$. For each $x \in O$, note $x \in B_d(x, \epsilon) \subseteq O$ for some $\epsilon > 0$. By hypothesis, there is some $\delta > 0$ such that $B_x = B_{d'}(x, \delta)$ is such that $x \in B_x \subseteq B_d(x, \epsilon) \subseteq O$. Then $O = \bigcup_{x \in O} B_x \in \Tau'$. ❦

Proof.

We begin by showing the Euclidean and square metrics generate the same topologies on $\bR^n$. Recalling that basis elements for the Euclidean metric (resp. square metric) topology take the form of balls (resp. congruent boxes), we seek to show that a box can be contained in a sphere and vice versa. For starters, fix the points $x, y \in \bR^n$, and let $\vec{y}$ denote the vector whose tip is located at $x$ and tail is located at $y$. The magnitude of $\vec{y}$, representing the Euclidean distance $d(x,y)$, is bounded below by its largest single coordinate projection $\rho(x,y)$, meaning the box spanned by $\vec{y}$ easily fits within a sphere of radius $d(x,y)$. Conversely, if we bound $\vec{y}$ inside an $n$-dimensional cube centered at $x$ whose faces are at a distance of $\rho(x,y)$, the Pythagorean theorem guarantees its magnitude cannot exceed the distance to the cube's furthest corner, yielding the upper bound $d(x,y) \le \sqrt{n}\rho(x,y)$. The statement, which can also be proven with basic algebra, is that

In any case, this implies $B_d(x, \epsilon) \subseteq B_\rho(x, \epsilon)$ for any $x \in X, \epsilon > 0$, and same for $B_\rho(x, \epsilon/\sqrt{n}) \subseteq B_d(x, \epsilon)$. Thus the topologies are finer than eachother, making them equal.

It suffices to prove that the square metric generates the product topology. Let $\Tau_\rho$ be the topology generated by the square metric, and let $\Tau_\times$ be the product topology on $\bR^n$. First, let $O \in \Tau_\rho$. For each $x \in O$, there exists some $\epsilon > 0$ such that $x \in B_\rho(x, \epsilon) \subseteq O$. But now note there is some box centered about $x$ whose sidelengths are, at most, $\epsilon$—this is spanned by basis elements of the product topology, and so $x \in (\text{the box}) \subseteq B_\rho(x, \epsilon)$. Thus $\Tau_\rho \subseteq \Tau_\times$.

Now suppose $O \in \Tau_\times$. For each $x \in O$, there is some $B = (a_1, b_1) \times \cdots \times (a_n, b_n)$ centered about $x$ such that $x \in B \subseteq O$. With $\epsilon = \min\{|a_i - b_i|: 1 \le i \le n\}$, we note $x \in B_\rho(x, \epsilon/2) \subseteq B$, completing the proof. ❦

Proof.

Let $\vb{x} = (x_\lambda)_{\lambda \in \Lambda} \in \bR^\Lambda$. Let $U = \prod_{\lambda \in \Lambda} U_\lambda$ be an element of the product topology's basis about $\vb{x}$, where $\lambda_1, \dots, \lambda_n$ are the indices for which $U_\lambda \ne \bR$.

Let $z \in U$. We seek to show there is some $\epsilon > 0$ such that $z \in B_{\bar{\rho}}(x, \epsilon) \subseteq U$. For each $\lambda_i$, choose $\epsilon_i > 0$ such that $B_{\bar{d}}(x, \epsilon_i) \subseteq U_{\lambda_i}$. Then let $\epsilon = \min\{\epsilon_1, \dots, \epsilon_n\}$.

On the other hand, let $x \in B_{\bar{\rho}}(x, \epsilon)$ for some $\epsilon > 0$. For each $\lambda \in \Lambda$, let $X_\lambda = (x_\lambda - \epsilon/2, x_\lambda + \epsilon / 2)$, then note $x \in \prod_{\lambda \in \Lambda} X_\lambda \subseteq B_{\bar{\rho}}(x, \epsilon)$. ❦

Proof.

$(\Longrightarrow)$: Suppose $f$ is continuous. Fix $x \in X$ and $\epsilon > 0$, and let $O = B_{d_Y}(f(x), \epsilon)$. Since $O$ is open, it follows that $f^{-1}(O)$ is open. Since $f^{-1}(O)$ is the union of basic, open sets in $X$, there is some $\delta > 0$ such that $B_{d_X}(x, \delta) \subseteq f^{-1}(O)$. Then, for any $a \in B_{d_X}(x, \delta)$, it follows that $f(a) \in O$.

$(\Longleftarrow)$: Suppose that, for any $x \in X$ and $\epsilon > 0$, there is some $\delta > 0$ such that

Let $O$ be open in $Y$—we show $f^{-1}(O)$ is open in $X$. Let $x \in f^{-1}(O)$. Let $\epsilon > 0$ be such that $B_{d_Y}(f(x), \epsilon) \subseteq O$. By hypothesis, there is some $\delta > 0$ such that $B_x = B_{d_X}(x, \delta)$, where $B_x \subseteq f^{-1}(O)$. Then $f^{-1}(O) = \bigcup_{x \in f^{-1}(O)} B_x$, which is open. ❦

Proof.

$(\Longrightarrow)$: Suppose $(x_n) \subseteq A$ is such that $x_n \to x$. Then, for any open neighborhood $O$ of $x$, there is some $N \in \bZ_+$ such that $x_n \in O$ for every $n \ge N$. Thus $O \cap A \ne \emptyset$ and so $x \in \bar{A}$.

$(\Longleftarrow)$: Suppose $X$ has some metric $d$. Fix $x \in \bar{A}$, and fix some arbitrary open neighborhood $O$ of $x$. By definition of the closure, $O \cap A \ne \emptyset$. Since $O$ is open in a metric space, there is some $\epsilon > 0$ such that $B(x, \epsilon) \subseteq O$ and $B(x, \epsilon) \cap A \ne \emptyset$. Then there is some $N \in \bZ_+$ such that $1/n < \epsilon$ for every $n \ge N$. Let $x_n = a$ for some $a \in A$ and all $n < N$, and otherwise let $x_n$ be some arbitrary element of $B(x, 1/n) \cap A$, which is guaranteed to exist. Then $(x_n) \subseteq A$ and $x_n \to x$. ❦

Proof.

$(\Longrightarrow)$: Suppose $f$ is continuous. Let $(x_n) \subseteq X$ be a sequence such that $x_n \to x$. Let $O$ be an open neighborhood of $f(x)$. Then note $f^{-1}(O)$ is open by $f$'s continuity, and $x \in f^{-1}(O)$. Thus there is some $N \in \bZ_+$ such that $x_n \in f^{-1}(O)$ for every $n \ge N$. Thus $f(x_n) \in O$ for every $n \ge N$, meaning $f(x_n) \to f(x)$.

$(\Longleftarrow)$: Suppose $X$ has some metric $d$. Suppose that, for every $(x_n) \subseteq X$ such that $x_n \to x$, it follows that $f(x_n) \to f(x)$. Let $A \subseteq X$—we will show $f$ is continuous by proving $f(\bar{A}) \subseteq \bar{f(A)}$. For starters, suppose $x \in \bar{A}$. Since $X$ is metrizable, $A$ inherits a metric subspace topology, and by the sequence lemma there is some $(x_n) \subseteq A$ such that $x_n \to x$. By hypothesis, we have $f(x_n) \to f(x)$, meaning $f(x) \in \bar{f(A)}$. Thus $f(\bar{A}) \subseteq \bar{f(A)}$, completing the proof. ❦

Proof.

Let $x \in X$ and $\epsilon > 0$. We need to find $\delta > 0$ such that

Since $f_n \to f$ is uniform, there is some $N \in \bZ_+$ such that $d(f(x), f_N(x)) < \epsilon/3$ for every $x \in X$. Then there is some $\delta > 0$ such that

by continuity of each $f_n$. Then, if $d(x, y) < \delta$, we have that

completing the proof. ❦

Solution.

$(\Longrightarrow)$: Suppose $f_n \to f$ uniformly. To show $f_n \to f$ in $(\bR^X, \bar{\rho})$, we choose some arbitrary open neighborhood $O$ of $f$ and show there is some $N \in \bZ_+$ such that $f_n \in O$ for every $n \ge N$. Note there is some $\epsilon > 0$ such that $f \in B(f, \epsilon) \subseteq O$. By $f_n \to f$ being uniform, there is some $N_1 \in \bZ_+$ such that $|f_n(x) - f(x)| < 1$ for every $n \ge N_1$. Then

with the second equality coming from the aforementioned fact. Again, by uniformity of $f_n \to f$, there is some $N > N_1 \in \bZ_+$ such that $\bar{\rho}(f_n, f) \le \epsilon/2$ for every $n \ge N$. Thus $f_n \in B(f, \epsilon) \subseteq O$ for every $n \ge N$, meaning $f_n \to f$ as elements of $\bR^X$ as desired.

$(\Longleftarrow)$: Suppose $f_n \to f$ as elements of $\bR^X$. Fix $\epsilon > 0$. We seek to show there is some $N \in \bZ_+$ such that

for every $x \in X$ and $n \ge N$. Let $\epsilon' = \min\{\epsilon/2, 0.5\}$. Then let $O = B(f, \epsilon')$ be an open set in $\bR^X$. There is some $N \in \bZ_+$ such that $f_n \in O$ for each $n \ge N$ by hypothesis. For any $n \ge N$, note that $\bar{\rho}(f_n, f) < \epsilon' < 1$. Thus $\bar{d}(f_n(x), f(x)) < \epsilon' < 1$ for every $x \in X$, and so $\bar{d}(f_n(x), f(x)) = d(f_n(x), f(x)) < \epsilon' < \epsilon$ for every $x \in X$ and $n \ge N$, meaning $(f_n) \to f$ uniformly.