Real Analysis: Metric Topology
Metric Spaces
Metrics
Pythagoras gave us the formula for the distance between two points on the plane. The earlier section on Euclidean Space laid out the notion of a $p$norm, which is a generalization of the Pythagorean distance formula from the origin to a point in in $\mathbb{R}^n$ that uses the $p$th power and root, rather than strictly the second. Here, we generalize the notion of distance even further by considering distances between arbitrary elements of arbitrary sets, rather than points in $\mathbb{R}^n$. A function that measures distance in this way is called a metric. If we can prove things in terms of metrics in general, then the results we come up with will be valid for any metric in particular.
A metric on a set $X$ is a function $d : X \times X \rightarrow \mathbb{R}$ such that for all $x, y, z \in X$:
 Identity: $d(x, x) = 0$
 Positivity: $d(x, y) > 0$ if and only if $x \neq y$
 Symmetry: $d(x, y) = d(y, x)$
 Triangle Inequality: $d(x, z) \leq d(x, y) + d(y, z)$
As you can verify, the Manhattan norm and Euclidean norm on the difference of two vectors both form metrics on $\mathbb{R}^n$.
Metric Spaces
The tuple $(X, d)$ of a set $X$ and a metric $d$ on $X$ is called a metric space. We often refer to the set $X$ itself as the metric space for brevity, as the metric $d$ is either inferred as a customary default, such as the Euclidean metric on $\mathbb{R}^n$, understood to be arbitrary, or stated clearly nearby. Thus, if you see $\mathbb{R}^2$ referred to as a metric space itself, or with the "usual metric," know that it implies the Euclidean metric.
Metric Subspaces and Metric Superspaces
Consider the metric spaces $(X, d)$ and $(Y, g)$. $(X, d)$ is a metric subspace of $(Y, g)$ and $(Y, g)$ is a metric superspace of $(X, d)$ if and only if $X$ is a subset of $Y$ and $d$ is a restriction of $g$. Recall that a restriction of a function is just the portion of that function that's left over when we limit its domain to a smaller subset. For example, if $d$ is the Euclidean metric on $\mathbb{R}$, then $(\mathbb{N}, d\mathbb{N})$, $(\mathbb{Z}, d\mathbb{Z})$, and $(\mathbb{Q}, d\mathbb{Q})$ are all metric subspaces of $(\mathbb{R}, d)$.
Problems
The discrete metric on any set $X$ is defined as follows:
$d(x, y) = \left\{ \begin{array}{ll} 0 & \text{if } x = y \\ 1 & \text{if } x \neq y \end{array} \right.$
Verify that the discrete metric is a metric.
The identity and positivity requirements are satisfied by definition. To show symmetry, if $x=y$, then $d(x, y) = d(x, x) = 0 = d(y, y) = d(y, x)$, and if $x \neq y$, then $d(x,y) = 1 = d(y,x)$. To show the triangle inequality, we proceed by cases.

If $x = y = z$, then $d(x, z) = 0 \leq d(x, y) + d(y, z) = 0$.

If $x=y\neq z$, then $d(x,z) = 1 \leq d(x, y) + d(y, z) = 0 + 1 = 1$.

If $x\neq y = z$, then $d(x,z) = 1 \leq d(x, y) + d(y, z) = 1 + 0 = 1$.

If $x = z \neq y$, then $d(x, z) = 0 \leq d(x, y) + d(y, z) = 1 + 1 = 2$.

If $x \neq y \neq z$, then $d(x,z) = 1 \leq d(x, y) + d(y, z) = 1 + 1 = 2$.

Show that the Manhattan norm of the difference between two points in $\mathbb{R}^n$ is a metric. Namely, show that
$$d(\textbf{x}, \textbf{y}) = \\textbf{x}  \textbf{y}\_1 = \sum\limits_{i=1}^n x_i  y_i$$
is a metric on $\mathbb{R}^n$.

Identity:
$ d_p(x, x) = \sum\limits_{i=1}^n x_i  x_i\\ d_p(x, x) = \sum\limits_{i=1}^n 0 \\ d_p(x, x) = 0 \\ $

Positivity: If $x \neq y$, then there is at least one pair of terms $x_k$ and $y_k$ such that $x_k \neq y_k$ for some $k \in 1, \ldots n$. Therefore $x_k  y_k \neq 0$, so $x_k  y_k > 0$. Since all the terms in the sum are nonnegative, we see that $0 < x_k  y_k \leq \sum\limits_{i=1}^n x_i  y_i = d(x, y)$.

Symmetry:
$ d(x, y) = \sum\limits_{i=1}^n x_i  y_i \\ d(x, y) = \sum\limits_{i=1}^n (x_i  y_i) \\ d(x, y) = \sum\limits_{i=1}^n (y_i + x_i) \\ d(x, y) = \sum\limits_{i=1}^n y_i  x_i \\ d(x, y) = d(y, x) \\ $

Triangle Inequality:
$ d(x, z) = \sum\limits_{i=1}^n x_i  z_i \\ d(x, z) = \sum\limits_{i=1}^n (x_i  y_i) + (y_i  z_i) \\ d(x, z) \leq \sum\limits_{i=1}^n x_i  y_i + y_i  z_i \\ d(x, z) \leq \sum\limits_{i=1}^n x_i  y_i + \sum\limits_{i=1}^n y_i  z_i \\ d(x, z) \leq d(x, y) + d(y, z) \\ $

If $d$ is a metric on $X$ and $c$ is a positive constant, show that $f(x, y) = cd(x, y)$ is also a metric on $X$.

Identity: $f(x, y) = cd(x, x) = c0 = 0$.

Positivity: $f(x, y) = cd(x, y) > 0$ when $x \neq y$ by the second ordering axiom.

Symmetry: $f(x, y) = cd(x, y) = cd(y, x) = f(y, x)$.

Triangle Inequality:
$ f(x, z) = cd(x, z) \\ f(x, z) \leq c(d(x, y) + d(y, z)) \\ f(x, z) \leq cd(x, y) + cd(y, z) \\ f(x, z) \leq f(x, y) + f(y, z)$

Let $(X, d)$ be a metric space. Show that $d(x, y)  d(y, z) \leq d(x, z)$.
By the triangle inequality, we see that $d(x, y) \leq d(x, z) + d(z, y)$. Then $d(x, y)  d(z, y) \leq d(x, z)$. By the same inequality, we see that $d(z, y) \leq d(x, z) + d(x, y)$, and likewise that $d(z, y)  d(x, y) \leq d(x, z)$. Since $(d(x, y)  d(z, y)) = d(z, y)  d(x, y)$, we conclude that $d(x, y)  d(z, y) = d(x, y)  d(y, z) \leq d(x, z)$.
Let $(X, d)$ and $(Y, f)$ be metric spaces. Determine whether $(X \times Y, g)$ is a metric space where $g(a, b) = d(a_x, b_x)f(a_y, b_y)$.
Consider the points $(a, b), (a, c) \in X \times Y$ where $b \neq c$. The two points are not equal, but $g((a, b), (a, c)) = d(a, a)f(b, c) = 0f(b, c) = 0$. Therefore $g$ does not fulfill the identity axiom for metrics, so $(X \times Y, g)$ is not a metric space.
Let $f : X \rightarrow \mathbb{R}$ be an injective function, and consider the function $d : X \times X \rightarrow \mathbb{R}$ where $d(x, y) = f(x)  f(y)$. Show that $d$ is a metric on $X$.
Identity: $d(x, x) = f(x)  f(x) = 0$.
Nonnegativity: Let $x, y \in X$ where $x \neq y$. Then $d(x, y) = f(x)  f(y) > 0$, since $f(x) \neq f(y)$ by injectivity.
Symmetry: Let $x, y \in X$. Then
$ d(x, y) = f(x)  f(y) \\ d(x, y) = (f(x)  f(y)) \\ d(x, y) = f(y)  f(x) \\ d(x, y) = d(y, x) $
Triangle Inequality: First assume $f(x)  f(z)$ is nonnegative. Then
$ d(x, z) = f(x)  f(z) \\ d(x, z) = (f(x)  f(z)) + (f(y)  f(y)) \\ d(x, z) = (f(x)  f(y)) + (f(y)  f(z)) \\ d(x, z) \leq f(x)  f(y) + f(y)  f(z) \\ d(x, z) \leq d(x, y) + d(y, z) \\ $
Conversely, if $f(x)  f(z)$ is negative, then
$ d(x, z) = f(z)  f(x) \\ d(x, z) = (f(z)  f(x)) + (f(y)  f(y)) \\ d(x, z) = (f(z)  f(y)) + (f(y)  f(x)) \\ d(x, z) \leq f(z)  f(y) + f(y)  f(x) \\ d(x, z) \leq f(x)  f(y) + f(y)  f(z) \\ d(x, z) \leq d(x, y) + d(y, z) \\ $
Show that the Euclidean norm of the difference between two points in $\mathbb{R}^n$ is a metric. Namely, show that
$$d(\textbf{x}, \textbf{y}) = \\textbf{x}  \textbf{y}\_2 = \sqrt{\sum\limits_{i=1}^n x_i  y_i^2}$$
is a metric.
Identity: $d(\textbf{x}, \textbf{x}) = \sqrt{\sum\limits_{i=1}^n x_i  x_i^2} = \sqrt{\sum\limits_{i=1}^n 0^2} = 0$.
Positivity: $d(\textbf{x}, \textbf{y}) = \sqrt{\sum\limits_{i=1}^n x_i  y_i^2} \geq 0$ because every term of the sum is a square and thus nonnegative, thus the sum is nonnegative, and so its square root is welldefined and nonnegative.
Symmetry: $d(\textbf{x}, \textbf{y}) = \sqrt{\sum\limits_{i=1}^n x_i  y_i^2} = \sqrt{\sum\limits_{i=1}^n y_i  x_i^2} = d(\textbf{y}, \textbf{x})$.
Triangle Inequality:
$ d(\textbf{x}, \textbf{z})^2 = \sum\limits_{i=1}^n (x_i  z_i)^2 \\ d(\textbf{x}, \textbf{z})^2 = \sum\limits_{i=1}^n (x_i  y_i) + (y_i  z_i)^2 \\ d(\textbf{x}, \textbf{z})^2 \leq \sum\limits_{i=1}^n (x_i  y_i)^2 + (y_i  z_i)^2 \\ d(\textbf{x}, \textbf{z})^2 \leq \sum\limits_{i=1}^n (x_i  y_i)^2 + \sum\limits_{i=1}^n(y_i  z_i)^2 \\ d(\textbf{x}, \textbf{z}) \leq \sqrt{\sum\limits_{i=1}^n (x_i  y_i)^2 + \sum\limits_{i=1}^n(y_i  z_i)^2} \\ d(\textbf{x}, \textbf{z}) \leq \sqrt{\sum\limits_{i=1}^n (x_i  y_i)^2} + \sqrt{\sum\limits_{i=1}^n(y_i  z_i)^2} \\ d(\textbf{x}, \textbf{z}) \leq d(\textbf{x}, \textbf{y}) + d(\textbf{y}, \textbf{z}) $
The second to last step follows from the algbraic property of the square root operation proven in the prior Algebra on $\mathbb{R}$ section.
Consider the unit circle $C_1 = \{ (x, y) \in \mathbb{R}^2 : \sqrt{x^2 + y^2} = 1 \}$, and consider $dC_1$ as the Euclidean distance between points in $C_1$. Now consider the following argument: $(C_1, dC_1)$ is not a subspace of $\mathbb{R}^2$ because the distance traveled between points goes outside $C_1$. Is this reasoning sound?
No. Remember that the definition of a subspace needs to meet two conditions. The first is that the set be a subset, and the second is that the function be a restriction. We can see that $C_1 \subset \mathbb{R}^2$ by definition, and $dC_1$ is also a restriction of the Euclidean metric by definition. Thus $(C_1, dC_1)$ is a metric subspace of $(\mathbb{R}^2, d)$. Whether we may visualize the distance between two points in $C_1$ as a line that goes outside of $C_1$ is irrelevant. More specifically, the fact that it may be impossible to construct a geometric path through the subspace between two points whose length matches the distance determined by the metric is not a requirement.
Show that if $d(a, b) < \varepsilon$ for all $\varepsilon > 0,$ then $a = b.$
Proof by contradiction. Assume $a \neq b.$ Then $d(a, b) > 0.$ Pick $\varepsilon = \frac{1}{2}d(a, b).$ Then $d(a, b) < \frac{1}{2}d(a, b),$ which means that $\frac{1}{2}d(a, b) < 0$ and thus $d(a, b) < 0.$ However, this is a contradiction, as $d$ is a metric which is always nonnegative. Therefore $a = b$ after all.