Topology: Nets and Points of Accumulation

Recall that a sequence $(x_n)$ in a topological space X converges to a in X if the function f : N* → X which takes $n\mapsto x_n, \infty\mapsto a$ is continuous at $\infty\in\mathbf{N}^*$ . Unrolling the definition, it means that for any open subset U of X containing a, the set $f^{-1}(U)\subseteq \mathbf{N}^*$ contains (N, ∞] for some N. In other words,

for any open subset $U\subseteq X$ containing a, there exists N such that when n > N, $x_n\in U$ .

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 $Y\subseteq X$ is a subspace and $(x_n)$ is a sequence in Y converging to $a\in X$ . Does it mean $a \in Y$ ?

Even in the case where X = R, this is not true for some subspaces, e.g. Y = (0, 1), where the sequence $x_n = 1/n$ 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 a point of accumulation for Y if for every open subset $U\subseteq X$ containing a, there exists $x\in U\cap Y$ , 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 $(x_n)$ is a sequence in Y converging to $a\in X$ , then $a\in Y$ .

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_n) is a sequence in Y converging to a; thus a is in Y.

(2) → (3) : let $a\in X-Y$ , 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. $U\subseteq X-Y$ and X–Y is open in X.

(3) → (1) : suppose (x_n) is a sequence in Y converging to a. If $a\in X-Y$ , which is open in X, then there exists N such that when n>N, x_n lies in 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. A directed set is a partially ordered set I such that for any $i, j\in I$ , there exists $k\in I$ such that $i\le k, j\le k$ . In words, this means that every finite subset has an upper bound.

A net in a topological space X (indexed by I) is a function f : I → X, written as $(x_i)_{i\in I}$ .

We say that a net $(x_i)$ has limit $a\in X$ if for any open subset $U\subseteq X$ containing a, there exists an index i such that for all j ≥ i, $x_j \in U.$

Now the following is true.

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

If $(x_i)$ is a net in Y which converges to $a\in X$ , then $a\in Y.$

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: $i \le j$ iff $U_i \supseteq U_j.$ Since a is a point of accumulation of Y, for each $i\in I$ we can pick $x_i \in U_i\cap Y, x_i \ne a.$ Then $(x_i)_{i\in I} \to a$ so $a\in Y.$

(2) → (3) : identical to earlier.

(3) → (1) : let $(x_i)$ be a net in Y converging to $a\in X-Y.$ Since X–Y is open in X, there exists index i such whenever j≥i, $x_j \in X-Y,$ 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 $I^* := I \cup \{\infty\}$ , where ∞ is just a dummy symbol. A basis for open subsets of I* is given by:

$P_i := \{j\in I: j\ge i\} \cup \{\infty\}$ , for various $i\in I.$

To check that this is a basis, we claim that $P_i \cap P_j = \cup_{k\ge i, k\ge j} P_k.$

Indeed, clearly RHS is contained in LHS since $k\ge i \implies P_k \subseteq P_i.$
Conversely, if $k\in P_i\cap P_j$ is an index, then k ≥ i and k ≥ j so $k\in P_k$ is found in the RHS. The only case left is $\infty \in P_i \cap P_j.$ 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 $(x_i) \to a$ is exactly the same as that $\lim_{i\to\infty} x_i = a$ 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 $(x_i) \to a$ gives $(g(x_i)) \to g(a).$

The proof follows from the proposition here.

Proposition 5. If $(x_i, y_i)$ is a net in X × Y, then $(x_i, y_i) \to (a, b) \in X\times Y$ if and only if $(x_i)\to a$ and $(y_i)\to b.$

Proof.

We have $(x_i, y_i) \to (a, b)$ if and only if the function I* → X × Y which takes $i\mapsto (x_i, y_i)$ and $\infty \mapsto (a, b)$ 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 $(x_i) \to a$ and $(y_i)\to b.$ ♦

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 $(x_n) \to a$ , we also have $(f(x_n)) \to f(a).$

If f : X → Y is a map of topological spaces, then f is continuous if and only if for any convergent net $(x_i) \to a$ , we also have $(f(x_i)) \to f(a).$

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 $x_n\in X$ such that $d_X(x_n, a) < \frac 1 n$ but $d_Y(f(x_n), f(a)) \ge \epsilon.$ Then $(x_n) \to a$ but $(f(x_n)) \not\to a.$

Second statement: if f is not continuous at a, there exists an open subset $V\subseteq Y,$ $f(a)\in V,$ such that $f^{-1}(V)$ 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: $i\le j$ iff $U_i\supseteq U_j.$ Since no U_i is contained in f^-1(V), we can pick $x_i \in U_i - f^{-1}(V)$ for each i. Then $(x_i)$ is a net approaching a but $(f(x_i))$ doesn’t approach f(a) since they lie outside V. ♦