Can a net become a sequence?

Question

I am beginning to learn about nets. While reading about the topic I thought of the following:

Let $(X,\tau)$ be a topological space and $(x_i)_{i \in I}$ be a net converging to $x$. Suppose $f:X \rightarrow \mathbb{R}$ is a map. I know that $(x_i)_{i \in I}$ is a net. But what happens if one considers $(f(x_i))_{i \in I}$. It does seem logical to assume that $(f(x_i))_{i \in I}$ is a net, similarly to how $(f(x_i))_{i \in I}$ would be a sequence if $(x_i) $ would be a sequence.

The thing is, while $(X,\tau)$ might not be first-countable, I know that $(\mathbb{R},|.|)$ is first-countable. So to work in $(\mathbb{R},|.|)$ with nets seems unreasonable to me (I may be wrong here, as mentioned still at the beginning of learning.) This has led to the question of what happens with $(f(x_i))_{i \in I}$ in this situation?

---

My thoughts on that Question: I first wrote down the essential definitions, see below.

Definition:(directed set)
A directed set is a pair $(I,\leq)$ such that
(1) $\forall i \in I: i\leq i$
(2) $\forall i,j,k \in I: i \leq j$ and $j \leq k \Rightarrow i \leq k$
(3) $\forall i,j \in I$ there exists $k \in I$ such that $i \leq k$ and $j \leq k$

Definition (net)
Let $X$ be a set. A net is a map $x:I \rightarrow X$, where $(I,\leq)$ is a directed set.

Definition (convergence of net)
Let $(X,\tau)$ be a topological space, $(x_i)_{i \in I}$ a net and $x \in X$. The net $(x_i)_{i \in I}$ converges to $x$, iff $\forall N \in \mathcal{N}(x) \exists i_0 \in I \forall i \in I: i_0 \leq i \Rightarrow x_i \in N$.

---

I know that $(\mathbb{N}, \leq )$ with the normal ordering relation of "less or equal" is a directed set. So if $(I, \preceq )$ is a directed set, then maybe there exists some map $p: I \rightarrow \mathbb{N}$ , $p(I)=K \subseteq \mathbb{N}$ such that $(f(x_k))_{k \in K}$ is a sequence.

Answer 1

The object $\big(f(x_i)\big)_{i\in I}$ is quite simply a net. This is perfectly reasonable, even if $f$ maps to $\mathbb{R}$.

Such nets routinely show up when considering the continuity of $f$: $f$ is continuous if and only if the net $\big(f(x_i)\big)_{i\in I}$ converges to $f(x)$ whenever $(x_i)_{i\in I}$ converges to $x$. So checking continuity means checking that all limits in $X$ are preserved by $f$. This gives the intuition why we need to check all nets $\big(f(x_i)\big)_{i\in I}$, not just sequences $\big(f(x_n)\big)_{n\in \mathbb{N}}$: if we only consider such sequences we can easily miss limits in $X$.

As an example from functional analysis: all locally convex spaces $V$ have their topology generated by their seminorms $p: V\to \mathbb{R}^+$. Still, many locally convex spaces are not sequential and so require using nets to check things like closure and continuity. Since the topology is determined by the seminorms, which map to $\mathbb{R}$, one very often considers nets in $\mathbb{R}$.

Finally, I will remark that for functions $g: Y\to X$, with $Y$ some sequential space (e.g. C1, $\mathbb{R}$), it is much less natural to use nets. In this case $g$ is continuous iff the sequence $\big(g(y_n)\big)_{n\in \mathbb{N}}$ converges to $g(y)$ whenever $(y_n)_{n\in \mathbb{N}}$ converges to $y$. In this case nets are not necessary, since we just want to check that $g$ preserves all limits in $Y$ and all limits in $Y$ are obtainable by sequences.

Answer 2

Can a net become a sequence?

No, unless it is already a sequence.

Let us forget convergence for the moment. A net in a set $X$ is a function $x : I \to X$ defined on a directed set $I$. If $I = \mathbb N$, then $x$ is called a sequence.

If you have a function $f : X \to Y$, you get a net $f \circ x :I \to Y$ defined again on $I$. Thus, if $I \ne \mathbb N$, then $f \circ x$ cannot be a sequence.

Now let $X, Y$ be topological spaces. Obviously, if $x$ converges to $\xi \in X$ and $f$ is continuous at $\xi$, then $f \circ x$ converges to $f(\xi) \in Y$.

That said, I think what you really want to know is this:

1. Do we need general nets in $\mathbb R$?

2. Isn't it completely sufficient to work with sequences in that case?

3. If we have a net $x$ in $\mathbb R$, can we "compress" it to a sequence withount losing information?

1. I suggest that you read my answer to What is the motivation for sequences to be defined on natural numbers? You will see that nets play a vital role in elementary calculus (Riemann integral).

2. It depends on what you want to do. In fact all topological properties of $\mathbb R$ can be expressed via sequences. This is true because $\mathbb R$ is first-countable. See again What is the motivation for sequences to be defined on natural numbers? However, working only with sequences is not expedient; one would renounce important concepts of approximating real numbers by systems more general than sequences.

3. You might have the idea that each net $x : I \to \mathbb R$ has a subsequence, i.e. a subnet defined on $\mathbb N$, and that it could be that each net can be replaced by the collection of its subsequences.

Concerning the concept of subnet see A simpler equivalent definition of subnet.

The existence of a subsequence of a net $x: I \to X$ means that there is a sequence $y: \mathbb N \to \mathbb R$ and a monotonic function $ \varphi : \mathbb N \to I$ such that

• $y = x \circ \varphi$
• The image of $\varphi$ is cofinal in $I$

If only the first condition is satisfied, let us use the phrase subordinated sequence (this is just ad-hoc terminology).

Lemma. A net $x$ has a subsequence if and only if $I$ contains of countable cofinal subset.

Proof. The "only if" part follows from the definition. For the "if" part let $N$ be a countable cofinal subset of $I$ (it may be finite). There exists a surjection $\psi : \mathbb N \to N$. Define $\varphi$ inductively as follows. Take $\varphi(1) = \psi(1)$. Assume that we have constructed $\varphi(1) \le \varphi(2) \le \ldots \le \varphi(n)$ such that $\varphi(i) \ge \psi(i)$. Then take any $\varphi(n+1) \ge \varphi(n), \psi(n+1)$. Then certainly $y = x \circ \varphi$ is a subordinated sequence. Given $\alpha \in I$, there exists $\nu \in N$ such that $\nu \ge \alpha$. Write $\nu = \psi(n)$. Then $\varphi(n) \ge \alpha$, i.e. the image of $\varphi$ is cofinal in $I$.

For some $I$ the existence of a countable cofinal subset is trivial. Examples are $I = \mathbb R$ with its usual order and $I =$ set of finite subsets of $\mathbb N$ ordered by inclusion.

In contrast, to find subordinated sequences is easy. For example, all constant maps $\varphi : \mathbb N \to I$ will do the job.

Concerning convergence: It is obvious that if a net $x$ converges to $\xi \in X$, then all subnets (in particular all subsequences) converge to $\xi$. Cofinality is essential here.

The convergence of a single subnet of course does not imply the convergence of the full net $x$.

Example. In the context of the Riemann integral we consider $I =$ set of all partitions of a closed interval $[a,b]$. This does not contain a countable cofinal subset, hence no net on $I$ has a subsequence.

In fact, let $N \subset I$ be any countable set. Then the set $P$ of all partition points occurring in the members of $N$ is countable. Take any $\tau \in [a,b] \setminus P$. Then there is no partition in $N$ refining the partition $(a, \tau, b)$.

But perhaps cofinality is not that important in the special case $X = \mathbb R$? Does the convergence of subordinated sequences tell us anything about the convergence of $x$?

No, we do need cofinality. Concerning subordinated sequences we only have this:

If $x$ converges to $\xi \in \mathbb R$, then there exists a subordinated sequence converging to $\xi$.

Let us construct $\varphi$ inductively. Pick any $\alpha_1 = \varphi(1)$ such that $x(\alpha_1) \in (\xi-1,\xi+1)$. Assume we have $\varphi(1) \le \varphi(2) \le \ldots \le \varphi(n)$ such that $x(\varphi(i)) \in (\xi-1/i,\xi+1/i)$. Since $x$ converges to $\xi$, there exists $\beta \in I$ such that $x(\gamma) \in (\xi-1/(n+1),\xi+1/(n+1))$ for $\gamma \ge \beta$. Now choose $\varphi(n+1) \ge \varphi(n), \beta$.

The converse is of course not true. All constant $\varphi : \mathbb N \to I$ produce subordinated sequences of $x$ which are convergent, but there is no reason why $x$ should be convergent.