Proofs in Topology

Metric Spaces

The pedagogical arc of mathematical analysis traditionally commences with the $\epsilon-\delta$ characterization of continuity. In the standard Euclidean setting of $\mathbb{R}$, we assert that a function f is continuous at $a \in \mathbb{R}$ if, for every $\epsilon > 0$, there exists $a \delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon$. This formulation relies fundamentally on the distance-based intuition of “closeness.” However, a rigorous examination reveals that the “closeness” of real numbers is merely a specific instantiation of a far more generalized structure.

By isolating the four fundamental properties of the absolute difference $|x - y|$:

• non-negativity,
• the identity of indiscernibles,
• symmetry, and
• the triangle inequality

we can abstract the concept of distance into a formal metric. This transition allows us to move beyond the real number line into arbitrary sets, where “distance” may be defined in various ways. Ultimately, we shall see that even the metric itself can be discarded in favor of a purely set-theoretic framework: the topological space.

Consider $X = \mathbb{R}$ with $d(x, y) = |2^x - 2^y|$. We verify its metric properties:

• Properties (1) and (3): Inherited directly from the properties of absolute value.
• Property (2):

If $d(x, y) = 0$, then $|2^x - 2^y| = 0$, implying $2^x = 2^y$.

Since the function $f(x) = 2^x$ is strictly monotonic and thus injective, we may take the logarithm to the base $2$ of both sides to conclude $x = y$.

• Property (4):

$$ d(x, z) = |2^x - 2^z| = |(2^x - 2^y) + (2^y - 2^z)| \leq |2^x - 2^y| + |2^y - 2^z| = d(x, y) + d(y, z) $$

The geometry of a space is dictated by the definition of its metric. We examine three fundamental examples:

The Euclidean Metric in $\mathbb{R}^2$

For $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$,

$$ d(P_1, P_2) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} $$

The Manhattan (Taxicab) Metric Defined as

$$ d(P_1, P_2) = |x_1 - x_2| + |y_1 - y_2| $$

This metric models distance constrained to a grid, such as urban blocks.

The Discrete Metric For any nonempty set $A$,

$$ d(x, y) = \begin{cases} 0, & \text{ if } x = y \\ 1, & \text{ if } x \neq y \\ \end{cases} $$

The introduction of a metric provides the necessary machinery to define neighborhoods through the construction of open spheres.

Open Sets in Metric Spaces

The “shape” of $S_r(a)$ varies by metric:

• Euclidean $\mathbb{R}$: $S_r(a)$ is the open interval $(a - r, a + r)$.
• Euclidean $\mathbb{R}^2$: $S_r(a)$ is the interior of a disk.
• Manhattan $\mathbb{R}^2$: $S_3(5, 4)$ is the interior of a “tilted square” with vertices at $(2, 4), (8, 4), (5, 1)$, and $(5, 7)$.

Proof

Let $x \in S_r(a)$, so $d(x, a) < r$. Define

$$r' = r - d(x, a) > 0$$

For any $y \in S_{r’}(x)$, we have $d(y, x) < r’$. By the triangle inequality:

$$d(y, a) \leq d(y, x) + d(x, a) < r' + d(x, a) = (r - d(x, a)) + d(x, a) = r$$

Thus $y \in S_r(a)$, proving $S_{r’}(x) \subseteq S_r(a)$.

Proof

Assume $O$ is open. For each $x \in O$, there exists $r_x > 0$ such that $S_{r_x}(x) \subseteq O$.

Then $\bigcup_{x \in O} S_{r_x}(x) \subseteq O$. Since $x \in S_{r_x}(x)$ for each $x$, it follows that $O \subseteq \bigcup_{x \in O} S_{r_x}(x)$. Thus $O = \bigcup_{x \in O} S_{r_x}(x)$.

If $O$ is a union of open spheres, any $x \in O$ belongs to some $S_r(a) \subseteq O$. Since $S_r(a)$ is open, there exists $S_{r’}(x) \subseteq S_r(a) \subseteq O$. Hence $O$ is open.

Proof

Let $x \in (S_r[a])^c$, meaning $d(x, a) > r$. Let $r^* = d(x, a) - r > 0$.

For any $y \in S_{r^*}(x)$, we have $d(y, x) < r^*.$ By the triangle inequality

$$d(x, a) \leq d(x, y) + d(y, a)$$

which implies:

$$d(y, a) \geq d(x, a) - d(x, y) > d(x, a) - r^* = r$$

Thus $y \in (S_r[a])^c$, proving the complement is open.

Continuity in Metric Spaces

Proof:

If $f$ is continuous and $O$ is open, then for any $x \in f^{-1}(O)$, $f(x) \in O$. There exists $S_\epsilon(f(x)) \subseteq O$.

By continuity, there exists $\delta$ such that

$$f(S_\delta(x)) \subseteq S_\epsilon(f(x)) \subseteq O$$

Thus $S_\delta(x) \subseteq f^{-1}(O)$, and $f^{-1}(O)$ is open. The converse follows by considering $O = S_\epsilon(f(a))$.

Topological Spaces

By stripping away the distance function, we identify the minimal conditions required to study continuity and convergence. “Topological spaces” allow us to generalize these concepts to environments where a metric may be irrelevant or entirely absent.

The members of $\tau$ are the open sets of the space. Consider the two extremes of this hierarchy:

The Trivial Topology

The coarsest possible structure, providing no internal resolution.

$$ \tau_1 = \{\emptyset, X\} $$

One must recognize that while all metric spaces are topological spaces, the converse is false. The topological axioms are far more permissive than the rigid requirements of a distance function.

Consider the set $X = \{a, b, c\}$ with the topology

$$ \tau = \{\emptyset, X, \{a\}, \{a, b\}, \{a, c\}\} $$

We can prove this space is not metrizable by contradiction.

Assume a metric $d$ exists that generates $\tau$. Let

$$ r = \min\{d(a, b), d(b, c)\} $$

Since $a, b, c$ are distinct, $r > 0$. The open sphere

$$ S_r(b) = \{x \in X : d(x, b) < r\} $$

would necessarily exclude $a$ (since $d(a,b) \ge r$) and $c$ (since $d(b,c) \ge r$). Thus, $S_r(b) = \{b\}$. However, $\{b\} \notin \tau$.

Since every open sphere in a metric space must be an open set in its topology, this contradiction proves that no such metric $d$ can exist.

The Hausdorff Property

In non-Hausdorff spaces, standard analytical concepts like the uniqueness of limits for sequences break down. Metric spaces, however, are shielded from such pathology:

Continuity in Topological Spaces

In the abstract setting, the definition of continuity is purely topological: $f: X \to Y$ is continuous if $f^{-1}(O)$ is open in $X$ for every open $O \subseteq Y$.

Behavior under Extremal Topologies

Discrete Topology

If $X$ has the discrete topology, every function $f: X \to Y$ is continuous because $f^{-1}(O)$, being a subset of $X$, is automatically open.

Trivial Topology

If $X$ has the trivial topology and $f$ is surjective, $f$ is continuous if and only if $Y$ also has the trivial topology.

If $Y$ possessed a non-trivial open set $O$, the surjectivity of $f$ ensures that $f^{-1}(O)$ is neither $\emptyset$ nor $X$.

Since only $\emptyset$ and $X$ are open in the trivial topology, $f$ would fail to be continuous. Surjectivity is the key here; without it, $f^{-1}(O)$ might still map to $\emptyset$, masking the non-triviality of $Y$.

Neighborhood Equivalence

The modern definition of continuity is functionally equivalent to the neighborhood-based definition of classical analysis.

Proof

If $f$ is continuous, for any $x$ and open $O$ containing $f(x)$, we set $U = f^{-1}(O)$. By definition, $U$ is open, contains $x$, and $f(U) = f(f^{-1}(O)) \subseteq O$ by Lemma 16.26.

Conversely, let $B$ be an open set in $Y$. For each $x \in f^{-1}(B)$, we have $f(x) \in B$.

By hypothesis, there is an open set $U_x$ containing $x$ such that $f(U_x) \subseteq B$, meaning $U_x \subseteq f^{-1}(B)$. Then $f^{-1}(B) = \bigcup_{x \in f^{-1}(B)} U_x$.

Since the union of open sets is open, $f^{-1}(B)$ is open, and $f$ is continuous.