Prove the following useful fact, which you may use later on this assignment. You do not need to submit this problem.
Suppose that $M$ is a set and $f : M \times M \to \R_{\geq0}$ is such that for all $x, y \in M$, $f(x, y) = 0$ if and only if $x = y$, and $f(x, y) = f(y, x)$. Prove that if $x, y \in M$, we have \begin{align*} f(x, x) &\leq f(x, y) + f(y, x) \\ f(x, y) &\leq f(x, y) + f(y, y) \\ f(x, y) &\leq f(x, x) + f(x, y) \\ f(x, x) &\leq f(x, x) + f(x, x). \end{align*} Consequently, to show $f$ is a metric, it suffices to check that the triangle inequality holds for triples of distinct points.
Recall that a function $f : X \to M$ is said to be injective or one-to-one if for every $x, y \in X$, $f(x) = f(y)$ implies $x = y$.
Suppose $(M, d)$ is a metric space, $X$ is a set, and $f : X \to M$. Define \begin{align*} d_f : X\times X &\longrightarrow \R_{\geq0} \\ (x, y) &\longmapsto d(f(x), f(y)). \end{align*} Prove that $d_f$ is a metric on $X$ if and only if $f$ is injective.
Let $M$ be a set. Given a subset $E \subseteq M$, we denote by $E^c$ the complement of $E$ in $M$: $E^c = M \setminus E = \set{x \in M \mid x \notin E}$. Note that this notation only makes sense when $M$ is clear.
Verify the following identities, where $(E_\alpha)$ is a collection of subsets of $M$.
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.
Let $S$ be a set. Call a function $d : S\times S \to \R_{\geq0}$ an anti-metric if for every $x, y, z \in S$:
Show that if $d$ is an anti-metric on $S$, then $S$ contains at most one element.
Suppose $M$ and $X$ are metric spaces. We say $X$ embeds into $M$ (and write $X \hookrightarrow M$) if there is a function $f : X \to M$ which preserves distance: that is, if for all $x, y \in X$ we have $d_X(x, y) = d_M(f(x), f(y))$. Such a function $f$ is said to be an isometry. Notice that the function $f$ in Problem 2 is an isometry from $(X, d_f)$ to $(M, d)$. Also notice that an isometry is necessarily injective.