Topology: Limits and Convergence

Following what we did for real analysis, we have the following definition of limits.

Definition of Limits. Let X, Y be topological spaces and a\in X. If  f : X-{a} → Y is a function, then we write \lim_{x\to a} f(x) = b \in Y if the function:

g:X\to Y,\ g(x) = \begin{cases} f(x), &\text{if } x\ne a, \\ b, &\text{if }x=a.\end{cases}

is continuous at x=a. In words, we say that f(x) approaches b as x approaches a.

Definition of Convergence. If X is a topological space, a sequence of elements x_n \in X is said to converge to a if the function f : N* – {∞} → X, f(n) = xn, approaches a as n approaches ∞. A sequence which has a limit is called a convergent sequence.

[ Refer here for the definition of N*. ]

sequence_in_topo

We shall show that this convergence is consistent with our earlier definition of convergence.

Proposition. Let (X, d) be a metric space. Then a sequence x_n\in X converges to a (in the above definition) if and only if:

  • for any ε>0, there exists N such that whenever n>N, d(x_n, a) < \epsilon.

Proof.

  • Suppose x_n\in\mathbf{R} converges to a.  Given any ε>0, N(a, ε) is an open ball containing a. By definition, if N* → X is defined by g(n)=xn, for n=1, 2, 3, … and g(∞)=a, then g is continuous at ∞. Thus U:=g^{-1}(N(a,\epsilon)) contains an open subset of N* containing ∞. So U contains some UN = {NN+1, N+2, … } and n>N\implies n\in U\implies x_n=g(n)\in N(a,\epsilon).
  • Conversely, assume the condition in the proposition holds; let’s prove that g is continuous at ∞. Now any open subset V\subseteq X containing a must contain an open ball N(a,\epsilon) \subseteq V for some ε>0. Then there exists N such that whenever n>N, d(x_n, a)<\epsilon \implies x_n \in N(a,\epsilon). Thus g^{-1}(V) contains the set UN+1 = {N+1, N+2, N+3, …, ∞} which is an open subset containing ∞. ♦

In addition to N*, let’s consider the extended real line \overline{\mathbf{R}} = \mathbf{R}\cup \{-\infty, \infty\}. Again we should think of the infinities as dummy symbols. Its topology is defined via the basis:

  • (a,b) for real ab;
  • (a,\infty] = (a, \infty) \cup \{\infty\} for real a;
  • [-\infty, b) = (-\infty, b)\cup \{-\infty\} for real b.

Then all prior limits can be expressed via the extended real line. For example:

Proposition.

  • If x_n is a sequence of real numbers, then \lim_{n\to\infty} x_n = \infty in the classical sense iff the function f:\mathbf{N}^*-\{\infty\}\to \overline{\mathbf{R}} which takes n\mapsto x_n tends to ∞ as n tends to ∞.
  • If f: \mathbf{R}\to\mathbf{R} is a function, then \lim_{x\to\infty} g(x) = L \in \mathbf{R} in the classical sense iff the function f:\overline{\mathbf{R}} - \{\infty\} \to \mathbf{R} tends to L as x→∞.

Proof.

We’ll prove (LHS) → (RHS) and leave the converse to the reader.

Let’s prove the first statement. Suppose LHS holds; we need to show g:\mathbf{N}^* \to \overline{\mathbf{R}} which takes n\mapsto x_n, \infty\mapsto \infty is continuous at ∞. Indeed, any open subset of \overline{\mathbf{R}} containing ∞ must contain some (L, ∞]. By classical definition, there exists N such that whenever n>N, we have xn>L. Then g^{-1}((L,\infty]) contains U_{N+1}\cup\{\infty\}, an open set containing ∞.

For the second statement, again assume LHS holds. Let g:\overline{\mathbf{R}} \to \mathbf{R} be the function which takes real x to f(x) and ∞ to L; our job is to prove continuity of g at ∞. Let V be an open subset of R containing L, which contains (L-ε, L+ε) for some ε>0. By classical definition, there exists M such that whenever x>Mf(x) lies in (L-ε, L+ε) and hence V. Thus, (M,\infty] \subseteq f^{-1}(V) is an open subset of \overline{\mathbf{R}} containing ∞, and g is continuous at ∞. ♦

The point we’re trying to say is this.

Summary. The various multi-case definitions of limits and convergence can all be subsumed under a single definition by considering various topological spaces.

blue-linBasic Properties of Limits

The following is basic.

Proposition. Let f : X-{a} → Y be a function where \lim_{x\to a} f(x) = b. If g : Y → Z is continuous at b, then gf : X-{a} → Z satisfies \lim_{x\to a} gf(x) = g(b).

Proof.

Define h:X\to Y and i:X\to Z via:

h(x)=\begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a,\end{cases}  and  i(x) = \begin{cases} gf(x), &\text{if } x\ne a,\\ g(b), &\text{if }x=a.\end{cases}

From the condition, we know that h is continuous at a. Since g is continuous at b=h(a), igh is also continuous at a. ♦

Corollary. If x_n \to a in the topological space X, and f : X → Y is continuous, then f(x_n)\to f(a) in Y.

Proof

Since x_n\to a, the function g:\mathbf{N}^* \to X which takes n\mapsto x_n has a limit \lim_{n\to\infty} g(n) = a. By the previous proposition, we get \lim_{n\to\infty} f(g(n)) = f(a) and so f(x_n)\to f(a). ♦

Next we shall approach the basic question: is the limit \lim_{x\to a} f(x) unique? Phrased in such generality, the answer is no, as we saw counter-examples even for the case where X is a subset of R.

So let’s first consider sequences.

Theorem. Let X be a topological space. Assume:

  • for any two distinct points x, y\in X, there exist open subsets U and V, x\in U, y\in V, U\cap V=\emptyset.

Then every sequence x_n \in X has at most one limit in X.  A topological space satisfying the above property is said to be Hausdorff.

hausdorff

Thus, the Hausdorff property is a sufficient condition for unique convergence, but it’s known that the condition is not necessary.

Proof.

Suppose ab are distinct limits of (x_n). Pick open subsets UV such that a\in U, b\in V, U\cap V=\emptyset. Then there exists MN such that (i) whenever n>M, we have x_n \in U, (ii) whenever n>N, we have x_n\in V. The two statements clearly contradict. ♦

One obvious source of Hausdorff topological spaces is via metric spaces, i.e. metric spaces are Hausdorff. Indeed, if xy are distinct points in a metric space X, then letting ε=d(xy)/2 > 0, we have N(x,\epsilon) \cap N(y,\epsilon) = \emptyset, for if z satisfies d(xz) < ε and d(yz) < ε, then d(xy) ≤ d(xz) + d(zy) < 2ε = d(xy), which is absurd. In particular, any convergent sequence in a metric space has a unique limit.

For uniqueness of general limits, obviously we have to care about the domain space X.

Theorem. Let f : X-{a} → Y be a map of topological spaces. Assume the following.

  • Y is Hausdorff.
  • The singleton set {a} is not open in X.

Then there’s at most one limit for \lim_{x\to a} f(x).

Proof.

Suppose b,c\in Y are distinct limits. Hence the functions g_1, g_2:X\to Y,

g_1(x) = \begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a\end{cases} and g_2(x)=\begin{cases} f(x), &\text{if }x\ne a,\\ c &\text{if }x=a\end{cases}

are continuous. Pick open subsets U and V of Y, such that b\in U, c\in V and U\cap V=\emptyset. Now g_1^{-1}(U) = f^{-1}(U)\cup\{a\} must contain an open subset which contains a. Same for g_2^{-1}(V) = f^{-1}(V) \cup\{a\}. But since f^{-1}(U)\cap f^{-1}(V) = f^{-1}(U\cap V)=\emptyset, the intersection of these two open subsets is {a}, which contradicts the second condition. ♦

Note: it’s easy to find a counter-example when {a} is open in X. In an earlier article, we saw the following:

kindacontinuous

Here X=(-\infty, -1)\cup \{0\}\cup (1, \infty) and fX-{0} → R is defined by f(x)=x. But because {0} is an open subset of Xf(0) can take any value without violating its continuity.

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