# Why is special about the reverse inclusion in the following example ?

Given a point $x$ in a topological space, let $N_x$ denote the set of all neighbourhoods containing $x$. Then $N_x$ is a directed set, where the direction is given by reverse inclusion, so that $S ≥ T$ if and only if $S$ is contained in $T$. For $S$ in $N_x$, let $x_S$ be a point in $S$. Then $(x_S)$ is a net. As $S$ increases with respect to $≥$, the points $x_S$ in the net are constrained to lie in decreasing neighbourhoods of $x$, so intuitively speaking, we are led to the idea that $x_S$ must tend towards $x$ in some sense.

I wonder if its only for fun :D

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The idea of a net is to generalize the notion of sequence convergence to something more powerful. Had they directed by inclusion, then moving along the net corresponds to a less tight focus around $x$, which is really going the wrong way if we want to talk about the $x_S$ tending toward $x$ in any reasonable sense. –  Cameron Buie May 24 '12 at 17:42
@CameronBuie but even i can use inclusion and formulate the same notion isn't it with respective change ? Am i missing something here ? –  Theorem May 24 '12 at 17:44
I've posted an example as an answer. Hopefully it helps clear things up, but let me know. –  Cameron Buie May 24 '12 at 18:02

## 2 Answers

The whole point of nets is to generalize the notion of sequences. In a metric space, for instance, the topology is completely determined by the convergent sequences: if you know which sequences converge, and what their limits are, you know in principle everything about the topology of the space. This is not true, however, of topological spaces in general. There are quite nice topological spaces with non-isolated points that are not the limit of any non-trivial sequence: they simply have in some sense too many open nbhds.

Example 1. For a very simple example, let $X=\Bbb R_{\ge0}$, the set of non-negative real numbers, and declare a set $V\subseteq X$ to be open iff either $0\notin V$, or $X\setminus V$ is countable. Then every nbhd of $0$ contains all but countably many positive real numbers, but no sequence of positive real numbers converges to $0$: if $\langle x_n:n\in\Bbb N\rangle$ is a sequence of positive real numbers, $X\setminus\{x_n:n\in\Bbb N\}$ is an open nbhd of $0$ that contains no term of the sequence. The only convergent sequences in this space are the ones that are constant from some point on. The point $0$ has such a variety of nbhds that it can always find one to ‘avoid’ any given sequence of positive real numbers.

Compare this with a very similar example in which sequences do determine the topology.

Example 2. Let $X=\Bbb Z$, and declare a set $V\subseteq\Bbb N$ to be open iff either $0\notin V$, or $X\setminus V$ is finite. Then every nbhd of $0$ contains all but finitely many positive integers, but this time there are sequences of positive integers converging to $0$. For example, the sequence $\langle n:n\in\Bbb Z^+\rangle$ of positive integers in their natural order converges to $0$: if $V$ is any nbhd of $0$, by definition $V$ contains all but finitely many terms of this sequence, so there is some $n_0$ such that $n\in V$ for all $n\ge n_0$.

Sequences are indexed by $\Bbb N$ or some ordered set with the same basic ‘shape’, like $\Bbb Z^+$. Intuitively, a sequence $\langle x_n:n\in\Bbb N\rangle$ converges to a point $x$ if its terms get closer to $x$ as $n$ increases and can be made to get as close as we like by taking $n$ large enough. This is formalized in the familiar way: for each open nbhd $V$ of $x$ there is an $n_V\in\Bbb N$ such that $x_n\in V$ whenever $n\ge n_V$. That is, no matter how small a nbhd of $x$ we choose as a target, a whole tail of the sequence hits that target. That works fine in the space of Example 2; it doesn’t work at all in the space of Example 1.

Now a sequence in $X$ is just a function from $\Bbb N$ to $X$. The problem with the space of Example 1 is that $\Bbb N$ isn’t big enough or complicated enough to capture that space’s notion of ‘nearness’. What’s needed is some kind of ordered set $\langle D,\le\rangle$ that is big enough and complicated enough to do the trick. Directed sets turn out to have the right kind of ordering: like $\Bbb N$ with its usual ordering, they have ‘tails’ that match up well with the intuitive idea of from some point on, and of course $\langle\Bbb N\le\rangle$ is itself a (very simple) directed set. Thus, if we define a net in $X$ to be a function from some directed set to $X$, we are indeed generalizing the notion of a sequence. The definition of convergence for nets is then copied directly from the definition for sequences:

Let $\langle D,\le\rangle$ be a directed set. A net $\langle x_d:d\in D\rangle$ in $X$ converges to a point $x$ iff for each open nbhd $V$ of $x$ there is a $d_V\in D$ such that $x_d\in V$ whenever $d\ge d_V$.

When $\langle D,\le\rangle=\langle \Bbb N,\le\rangle$, this is just the usual definition of sequential convergence.

Now, what would be a non-sequential example of a net converging to a point? What, for instance, might a net converging to $0$ in the space of Example 1 look like?

One natural candidate for a suitable directed set $-$ one that’s directly related to what happens in nbhds of $0$ $-$ is $\mathscr{N}_0$, the set of open nbhds of $0$. How should we choose the point $x_V$ corresponding to $V\in\mathscr{N}_0$? Since we want to get an indexed family of points that somehow get close to $0$, it makes sense to pick $x_V\in V$; $\langle x_V:V\in\mathscr{N}_0\rangle$ will be our net. That way we can at least be sure that every nbhd of $0$ contains a point of the net, which is surely a minimum requirement if the net is to converge to $0$.

Of course, $\mathscr{N}_0$ isn’t a directed set until we give it a suitable partial order. Since $\mathscr{N}_0$ is a family of sets, the natural candidates are $\subseteq$ and $\supseteq$. And in fact both $\langle\mathscr{N}_0\subseteq\rangle$ and $\langle\mathscr{N}_0\supseteq\rangle$ are directed sets: for any $U,V\in\mathscr{N}_0$, $U\cap V\in\mathscr{N}_0$, so $\mathscr{N}_0$ is directed by $\subseteq$, and $U\cup V\in\mathscr{N}_0$, so $\mathscr{N}_0$ is also directed by $\supseteq$. Let’s see how each works out.

$\langle\mathscr{N}_0,\subseteq\rangle$: $\langle x_V:V\in\mathscr{N}_0\rangle$ converges to $0$ iff for each open nbhd $U$ of $0$ there is a $V_U\in\mathscr{N}_0$ such that $x_V\in U$ whenever $V\subseteq V_U$.

Suppose that $U$ is an open nbhd of $0$; what would $V_U$ have to be? All we know about $x_V$ is that it’s in $V$; what would guarantee that it’s also in $U$? That’s easy: if $x_V\in V\subseteq U$, then obviously $x_V\in U$. If we take $V_U=U$, it will indeed be the case that $x_V\in U$ whenever $V\subseteq V_U$, and the net will converge to $0$.

$\langle\mathscr{N}_0,\supseteq\rangle$: $\langle x_V:V\in\mathscr{N}_0\rangle$ converges to $0$ iff for each open nbhd $U$ of $0$ there is a $V_U\in\mathscr{N}_0$ such that $x_V\in U$ whenever $V\supseteq V_U$.

Again suppose that $U$ is an open nbhd of $0$. How can we choose $V_U\in\mathscr{N}_0$ so that $x_V\in U$ whenever $V\supseteq V_U$? We can’t, unless $U=X$. As before, all we know about $x_V$ is that it’s in $V$. If $V\supsetneqq U$, we have no way to ensure that $x_V\in U$. The net cannot be shown to converge to $0$ (or to anything else). This choice of directed set simply isn’t useful.

The reason for this difference can be seen by looking at the tails of the two nets. Fix some $U\in\mathscr{N}_0$. The $U$-tail of the first net is $\{x_V:V\subseteq U\text{ and }V\in\mathscr{N}_0\}$, which is a subset of $U$. This tail could be described as being $U$-close to $0$, just as a suitable tail of a convergent sequence in a metric space can be described as being $\epsilon$-close to the limit of the sequence.

The $U$-tail of the second net, however, is $\{x_V:V\supseteq U\text{ and }V\in\mathscr{N}_0\}$. Points in this set could be scattered all over $X$ for all we know; there’s certainly nothing tying them closely to $0$. This is nothing like the behavior of tails of convergent sequences.

Added: Here’s a detailed proof that the sequence $\langle 1,2,3,\dots\rangle$ converges to $0$ in Example 2. To show this, we must show that if $V$ is any open nbhd of $0$, there is a positive integer $n_V$ such that $n\in V$ whenever $n\ge n_V$: some tail of the sequence is contained in $V$. So let $V$ be any open nbhd of $0$. By the definition of the topology on $X$, $X\setminus V$ is a finite set, say $X\setminus V=\{k_1,k_2,\dots,k_m\}$. Since $0\in V$, these numbers $k_1,\dots,k_m$ are positive integers, and we might as well arrange them in increasing order, $k_1<k_2<\dots<k_m$. Now let $n_V=k_m+1$. If $n\ge n_V=k_m+1$, then $$n>k_m>k_{m-1}>\dots>k_2>k_1\;,$$ and therefore $n\notin\{k_1,k_2,\dots,k_m\}$. That is, if $n\ge n_V$, then $n\notin X\setminus V=\{k_1,k_2,\dots,k_m\}$, and therefore $n\in V$. Since $V$ was an arbitrary open nbhd of $0$, we’ve now shown that for each open nbhd $V$ of $0$ there is a positive integer $n_V$ such that $n\in V$ whenever $n\ge n_V$. By definition, therefore, the sequence $\langle 1,2,3,\dots\rangle$ converges to $0$ in $X$.

It does not converge to any other point of $X$, however. I’ll illustrate this with a specific example: it does not converge to $17$. Remember, every point of $X$ except $0$ is an isolated point, so $\{17\}$ is an open nbhd of $17$. Is there any positive integer $m$ such that $n\in\{17\}$ whenever $n\ge m$? Clearly not: that would mean that every $n\ge m$ was equal to $17$, which is absurd.

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In the first example you said that $X$ \ $x_n$ is open nbd of $0$. I have a doubt , is $X$ \ $x_n$ countable to say that it is open nbd of $0$?? –  Theorem May 25 '12 at 7:49
@Ananda: Let $V=X\setminus\{x_n:n\in\Bbb N\}$; $V$ is open, becaues $X\setminus V=\{x_n:n\in\Bbb N\}$ is countable. A set containing $0$ is open if it complement is countable, not if the set itself is countable. –  Brian M. Scott May 25 '12 at 7:52
Can you explain me the sequence $x_n=n$ under this topology , how would it converge ? –  Theorem May 25 '12 at 8:13
@Ananda: In Example 2? Let $V$ be any nbhd of $0$. Then $X\setminus V$ is finite, say $X\setminus V=\{k_1,\dots,k_m\}$, where the $k_i$ are positive integers and $k_1<\dots<k_m$. Then $x_n\in V$ for every $n>k_m$. Thus, for every open nbhd $V$ of $0$ there is an $n_V$ ($k_m$ in my example) such that $x_n\in V$ for every $n>n_V$, and by definition of convergence, $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $0$. –  Brian M. Scott May 25 '12 at 8:18
M.Scott: i am not able to comprehend the first example :( . I don't have intuition to see how actually things work with complement topology. Can you suggest me something to understand it completely ! –  Theorem May 25 '12 at 8:54

Consider the topological space $X$ as the real line, with the following as the open sets: (i) $\emptyset,X$, (ii) $(-n,n)$ for all positive integers $n$, and (iii) $(-1/n,1/n)$ for all positive integers $n$. Then in fact, the neighborhoods of $x=0$ can be ordered (not just preordered) by inclusion, and in a way isomorphic with the natural order of $\mathbb{Z}$ with $+\infty$ included in its natural place at the top.

Now, if we say that $S\geq T$ iff $S\subseteq T$, and choose $x_S\in S$ for each neighborhood $S$ of $0$, then we see that as $S$ increases, $x_S$ tend closer to $0$--in the exact same way that one would expect, in fact.

On the other hand, if we say $S\geq T$ iff $S\supseteq T$, then as $S$ increases, it is entirely feasible that our points $x_S$ are getting arbitrarily far from $0$.

The idea of net convergence is that no matter how small the neighborhood of the point, there is some index sufficiently large (in the preorder) so that for all indices larger than that, the indexed points of the net lie in the neighborhood. In this particular instance, trying to preorder by inclusion cannot guarantee any such thing (though it could, of course, still happen). Does that help visualize the reason behind the choice?

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