KindiakMath

Converging Confusions

Topology was developed, in many ways, to generalise analysis and hence apply convergence-related notions to other mathematical objects.

Let’s first define convergence in the context of topological spaces, which largely resembles the \epsilon–\delta definition for the real numbers, but generalised into neighbourhoods.

Definition 1. Let (K, \mathcal T) be a topological space.

  • We call the elements of \mathcal T open sets.
  • For any x \in K, we call the open set V a neighbourhood of x if x \in V.
  • A K-sequence is a function \{x_n\} \equiv x \in \mathcal F(\mathbb N, K) \equiv K^\omega.

We say that the K-sequence x_n converges to x \in K if for any neighbourhood V of x, there exists N \in \mathbb N such that n \geq N implies x_n \in V.

In the context K = \mathbb R, letting V = (x-\epsilon, x+\epsilon) yields the usual \epsilon–\delta definition for limits. Furthermore, if the sequence \{ x_n \} converges, its limit is unique.

Remarkably, this property fails in general.

Lemma 1. For any set K, the collection K_{\mathrm{trivial}} := \{\emptyset, K\} forms a topology on K, called the trivial topology on K.

Example 1. For the topological space (\{0, 1\}, \{ 0, 1 \}_{\mathrm{trivial}}), the sequence x_n = 0 has both 0 and 1 as limits.

What could have failed? Let’s perhaps analyse a standard proof of the uniqueness of limits in the case of \mathbb R, recast in topological language. Suppose \{x_n\} is a real-valued sequence that converges to the limits x and y. If x,y are distinct, then \epsilon := d(x,y) > 0.

We leave it as an exercise to verify that (x-\epsilon/2,x+\epsilon/2) and (y-\epsilon/2,y+\epsilon/2) are non-overlapping neighbourhoods of the two limits. If both x and y are limits, then there exists some sufficiently large N such that for any n > N,

x_n \in (x-\epsilon/2,x+\epsilon/2) \cap (y-\epsilon/2,y+\epsilon/2) = \emptyset.

In particular, x_{N+1} \in \emptyset, a contradiction.

The key property we used is that for any two distinct points x,y, there exist disjoint neighbourhoods containing each element. In this case, we call the topological space Hausdorff.

Definition 2. A topological space K is Hausdorff if for any distinct x, y \in K, there exist disjoint neighbourhoods U,V of x,y respectively.

Example 2. For any topological space (K, \mathcal T), we say that K is metrizable if there exists a metric d on K such that \mathcal T = \mathcal T_d. Every metric space is obviously metrizable. Then every metrizable space is Hausdorff.

Proof. Fix distinct elements x, y \in K. Define r := d(x, y)/2 > 0. Then the neighbourhoods U = B_d(x, r) and V = B_d(y, r) are disjoint.

Theorem 1. Let K be a Hausdorff space. Then for any K-sequence \{x_n\}, if x_n converges to x,y \in K, then x = y, in which case we can write \lim_{n \to \infty}x_n = x without ambiguity.

Proof. Our proof is almost identical to that of our prior discussion. Nevertheless, we include a proof for completeness. Suppose instead that x,y are distinct. Since K is Hausdorff, there exist disjoint neighbourhoods U,V of x,y respectively. Since both x and y are limits, then there exists some sufficiently large N such that for any n > N,

x_n \in U \cap V = \emptyset.

In particular, x_{N+1} \in \emptyset, a contradiction.

Another property pertaining real numbers is that any function f : \mathbb R \to \mathbb R is continuous if and only if it is sequentially continuous. Let’s investigate this property in the context of topological spaces.

Definition 3. Let K, L be topological spaces. We say that f : K \to L is sequentially continuous if x_n \to x implies f(x_n) \to f(x). Here, we don’t assume that K,L are Hausdorff.

Theorem 2. Let K, L be topological spaces, f : K \to L a function.

  • If f is continuous, then f is sequentially continuous.
  • The converse holds if K is metrizable.

Proof. Suppose f is continuous. Fix x_n \to x. Let V be any neighbourhood of f(x). Since f is continuous, f^{-1}(V) is a neighbourhood of x. Since x_n \to x, there exists N \in \mathbb N such that f(x_n) \in V \iff x_n \in f^{-1}(V) for n \geq N, establishing continuity.

Suppose K is metrizable with a metric d. Let f : K \to L be sequentially continuous. Suppose f is not continuous. Then there exists x \in K and a neighbourhood V of f(x) such that f^{-1}(V) is not open. Since x \in f^{-1}(V), for any neighbourhood U of x, U \not\subseteq f^{-1}(V). This means U \cap K \backslash f^{-1}(V) \neq \emptyset.

Now, we observe that for any n \in \mathbb N, B_d(x, 1/n) is a neighbourhood of x. Hence, there exists x_n \in B_d(x, 1/n) \cap K \backslash f^{-1}(V) \subseteq K \backslash f^{-1}(V) . Furthermore, x_n \to x. By sequential continuity, f(x_n) \to f(x). Hence, there exists N \in \mathbb N such that f(x_n) \in V \iff x_n \in f^{-1}(V) for n \geq N. In particular,

\begin{aligned} x_N &\in f^{-1}(V) \cap K \backslash f^{-1}(V) = \emptyset, \end{aligned}

a contradiction.

Once again, without metrizability, bad things can happen.

Example 3. Define the co-countable topology \mathcal T_1 on \mathbb R via

\mathcal T_1 := \{K \subseteq \mathbb R : K = \emptyset\ \lor\ \mathbb R \backslash K\ \text{is countable}\}.

Define the discrete topology \mathcal T_2 := \mathcal P(\mathbb R). Then there exists a sequentially continuous function f : (\mathbb R, \mathcal T_1) \to (\mathbb R, \mathcal T_2) that is not continuous.

Proof. We leave it as an exercise to verify that \mathcal T_1 is not Hausdorff, and thus not metrizable. Let f = \mathrm{id}_{\mathbb R}. Suppose x_n \to x. If we can prove that x_n = x for sufficiently large n, then f(x_n) = x_n = x = f(x) for sufficiently large n, and f is indeed sequentially continuous.

To that end, consider the set K :=\{x_n : n \in \mathbb N\} \backslash \{x\}, which is countable. Defining L := \mathbb R \backslash K, we have \mathbb R \backslash L = \mathbb R \backslash (\mathbb R \backslash K) = K, which is countable, so that L \in \mathcal T_1. Since x \in L, L contains x. Since x_n \to x, there exists N \in \mathbb N such that x_n \in L for n \geq N. If x_n \neq x, then x \in K implies x \notin L, a contradiction. Hence, x_n = x, as required.

To prove that f is not continuous, consider the neighbourhood V := \{f(x)\} \in \mathcal T_2 \backslash \mathcal T_1 of f(x) = x. Then f^{-1}(V) = \{f(x)\} = \{x\} \notin \mathcal T_1. Hence, f is not continuous.

What’s interesting about Theorem 2 is that metrizability is not the most crucial component of the converse result. What we really used is that for any x \in K, there exists a countable collection \{B_d(x,1/n) : n \in \mathbb N\} of neighbourhoods of x.

Corollary 1. Suppose K is first-countable i.e. for any x \in K, there exist a countable collection \{U_n : n \in \mathbb N\} of neighbourhoods of x. Then f : K \to L is continuous if and only if it is sequentially continuous.

It is clear that the topological notions of analysis results are far more general, and of independent interest. The real-analytic conclusions arise due to the special topological properties satisfied by metric spaces, or metrizable spaces as a marginal generalisation, like first-countability in the context of Corollary 1.

We have many more ideas to revisit, like the cornerstone ideas of the extreme value theorem and the intermediate value theorem. These we will recast in topological flavours in future posts.

—Joel Kindiak, 28 Mar 25, 2302H

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