Recall that a sequence in a topological space *X* converges to *a* in *X* if the function *f* : **N*** → *X* which takes is continuous at . Unrolling the definition, it means that for any open subset *U* of *X* containing *a*, the set contains (*N*, ∞] for some *N*. In other words,

- for any open subset containing a, there exists
*N*such that when*n*>*N*, .

We also saw that if *X* is Hausdorff, then the limit *a* is unique, but we won’t make that assumption now.

The key question here is:

Suppose is a subspace and is a sequence in Y converging to . Does it mean ?

Even in the case where *X* = **R**, this is not true for some subspaces, e.g. *Y* = (0, 1), where the sequence converges to 0 which lies outside *Y*. In an earlier article, we saw that this is related to the requirement that *Y* is closed in *X*. Let’s attempt to reproduce that result for topological spaces.

Definition. Let a be a point in topological space X. We say it is apoint of accumulationfor Y if for every open subset containing a, there exists , x≠a.

## The Case of Metric Spaces

Let’s examine this definition for a metric space *X*. In this case, *a* is a point of accumulation for *Y* iff for each ε>0, the intersection *N*(*a*, ε) ∩ *Y* contains some point other than *a*. Indeed, (LHS) → (RHS) follows from the fact that *N*(*a*, ε) is an open subset containing *a*, while (RHS) → (LHS) follows from the fact that every open subset containing *a* must contain an open ball *N*(*a*, ε).

**Example**

Suppose *X* = **R** and *Y* is the set of rational numbers 0 < *r* < 1. Then any real number in [0, 1] is a point of accumulation for *Y* since for any ε>0, the open ball *N*(*r*, ε) must contain a rational number. [ The full rigourous argument needs a bit more care, but we won’t harp on this any more. ]

This result is reminiscent of an earlier one.

Theorem 1. In a metric space X and subspace Y, the following are equivalent.

- If is a sequence in Y converging to , then .
- Y contains all its points of accumulation.
- Y is closed in X.

**Proof.**

(1) → (2) : let *a* be a point of accumulation of *Y*. For each *n* = 1, 2, …, *N*(*a*, 1/*n*) ∩ *Y* contains a point *x _{n}* ≠

*a*. Then (

*x*) is a sequence in

_{n}*Y*converging to

*a*; thus

*a*is in

*Y*.

(2) → (3) : let , so it is not a point of accumulation of *Y*. By definition, there exists an open subset *U* containing *a* such that *U* ∩ *Y* has no point other than possibly *a*. But *a* is not in *Y* anyway, so *U* ∩ *Y* is empty, i.e. and *X*–*Y* is open in *X*.

(3) → (1) : suppose (*x _{n}*) is a sequence in

*Y*converging to

*a*. If , which is open in

*X*, then there exists

*N*such that when

*n*>

*N*,

*x*lies in

_{n}*X*–

*Y*, which is absurd. ♦

## The Case of Topological Spaces

We will see later that in the case of topological spaces, (2) and (3) are still equivalent, but (1) fails. The problem lies in the definition of sequences, which is essentially a function *f* : **N*** – {∞} → *X*, and the space **N*** is given unnecessary prominence. To overcome this one needs a more general form of sequences for topological spaces.

Definition. Adirected setis a partially ordered set I such that for any , there exists such that . In words, this means that every finite subset has an upper bound.A

netin a topological space X (indexed by I) is a function f : I → X, written as .We say that a net has

limitif for any open subset containing a, there exists an index i such that for all j ≥ i,

Now the following is true.

Theorem 2. Let Y be a subspace of a topological space X. The following are equivalent.

- If is a net in Y which converges to , then
- Y contains all its points of accumulation.
- Y is closed in X.

**Proof**

(1) → (2) : suppose *a* is a point of accumulation of *Y*. Let the index set *I* correspond to the collection of open subsets *U* containing *a*, ordered by reverse inclusion: iff Since *a* is a point of accumulation of *Y*, for each we can pick Then so

(2) → (3) : identical to earlier.

(3) → (1) : let be a net in *Y* converging to Since *X*–*Y* is open in *X*, there exists index *i* such whenever *j*≥*i*, which is a contradiction. ♦

## Interlude: Alternate Look at Nets

At this point, the reader may wonder why we need *I* to be a directed set, i.e. wouldn’t an ordinary partially ordered set (poset) suffice? Indeed, the above theorem seems to work even when *I* is just a poset.

To answer this question, we provide an alternate definition for nets and limits. Given a directed set *I*, let , where ∞ is just a dummy symbol. A basis for open subsets of *I** is given by:

- , for various

To check that this is a basis, we claim that

- Indeed, clearly RHS is contained in LHS since
- Conversely, if is an index, then
*k*≥*i*and*k*≥*j*so is found in the RHS. The only case left is*Since I is a directed set, there exists k≥i and k≥j*. Thus, the RHS is a non-empty union and ∞ is in the RHS. This proves that LHS is contained in RHS.

Thus, the fact that *I* is a directed set is necessary for *P _{i}*‘s to form a basis. Otherwise, they’d only form a subbasis in which case it’s possible for {∞} to be open in

*I**.

Next, the condition that a net is exactly the same as that where the limit was defined in the previous article. Since {∞} is not open in *I**, we conclude:

Proposition 3. If X is Hausdorff, a converging net in X has a unique limit.

Finally, we have the following.

Proposition 4. If g : X → Y is a continuous map of topological spaces, then every convergent net gives

The proof follows from the proposition here.

Proposition 5. If is a net in X × Y, then if and only if and

**Proof**.

We have if and only if the function *I** → *X* × *Y* which takes and is continuous. Now we use the following property, whose proof is left as an exercise.

- If
*X*,*Y*,*Z*are topological spaces, then*f*:*Z*→*X*×*Y*is continuous if and only if composing it with the projection maps*X*×*Y*→*X*and*X*×*Y*→*Y*, we get continuous maps*Z*→*X*and*Z*→*Y*.

So the above function *I** → *X* × *Y* is continuous if and only if composing with the projection maps give continuous functions *I** → *X* and *I** → *Y*. And this is true if and only if and ♦

## One Final Result

Another sign that nets are a good way to generalise sequences:

Theorem 6.If f : X → Y is a map of metric spaces, then f is continuous if and only if for any convergent sequence , we also have

If f : X → Y is a map of topological spaces, then f is continuous if and only if for any convergent net , we also have

**Proof**.

The forward direction in both cases has been proven in proposition 4 above and here. For the converse:

First statement: if *f* is not continuous at *a*, then there exists ε>0 such that for any δ>0, we can find *x* in *X* such that *d _{X}*(

*x*,

*a*) < δ but

*d*(

_{Y}*f*(

*x*),

*f*(

*a*)) ≥ ε. Letting δ = 1/

*n*for

*n*=1, 2, 3, … we get a sequence such that but Then but

Second statement: if *f* is not continuous at *a*, there exists an open subset such that doesn’t contain an open subset of *U* containing *a*. As before, let the index set *I* correspond to the collection of open subsets of *X* containing *a*, ordered by reverse inclusion: iff Since no *U _{i}* is contained in

*f*

^{-1}(

*V*), we can pick for each

*i*. Then is a net approaching

*a*but doesn’t approach

*f*(

*a*) since they lie outside

*V*. ♦