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

Definition of Limits. Let X, Y be topological spaces and . If f : X-{a} → Y is a function, then we write if the function: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 is said toconvergeto a if the function f :N* – {∞} → X, f(n) = x_{n}, approaches a as n approaches ∞. A sequence which has a limit is called aconvergent sequence.[ Refer here for the definition of

N*. ]

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 converges to a (in the above definition) if and only if:

- for any ε>0, there exists N such that whenever n>N, .

**Proof.**

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

In addition to **N***, let’s consider the **extended real line** Again we should think of the infinities as dummy symbols. Its topology is defined via the basis:

- for real
*a*<*b*; - for real
*a*; - for real
*b*.

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

Proposition.

- If is a sequence of real numbers, then in the classical sense iff the function which takes tends to ∞ as
ntends to ∞.- If is a function, then in the classical sense iff the function 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 which takes is continuous at ∞. Indeed, any open subset of containing ∞ must contain some (*L*, ∞]. By classical definition, there exists *N* such that whenever *n*>*N*, we have *x _{n}*>

*L*. Then contains , an open set containing ∞.

For the second statement, again assume LHS holds. Let 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*>*M*, *f*(*x*) lies in (*L*-ε, *L*+ε) and hence *V*. Thus, is an open subset of 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.

## Basic Properties of Limits

Proposition. Let f : X-{a} → Y be a function where . If g : Y → Z is continuous at b, then gf : X-{a} → Z satisfies .

**Proof**.

Define and via:

and

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

Corollary. If in the topological space X, and f : X → Y is continuous, then in Y.

**Proof**

Since , the function which takes has a limit . By the previous proposition, we get and so ♦

Next we shall approach the basic question: is the limit 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 , there exist open subsets U and V, , , .
Then every sequence has at most one limit in X. A topological space satisfying the above property is said to be

Hausdorff.

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

**Proof**.

Suppose *a*, *b* are distinct limits of . Pick open subsets *U*, *V* such that , , . Then there exists *M*, *N* such that (i) whenever *n*>*M*, we have , (ii) whenever *n*>*N*, we have . The two statements clearly contradict. ♦

One obvious source of Hausdorff topological spaces is via metric spaces, i.e. *metric spaces are Hausdorff*. Indeed, if *x*, *y* are distinct points in a metric space *X*, then letting ε=*d*(*x*, *y*)/2 > 0, we have , for if *z* satisfies *d*(*x*, *z*) < ε and *d*(*y*, *z*) < ε, then *d*(*x*, *y*) ≤ *d*(*x*, *z*) + *d*(*z*, *y*) < 2ε = *d*(*x*, *y*), 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 .

**Proof**.

Suppose are distinct limits. Hence the functions ,

and

are continuous. Pick open subsets *U* and *V* of *Y*, such that , and . Now must contain an open subset which contains *a*. Same for . But since , 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:

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