Let's say I have a topological space $(X,\tau)$ and a sequence $\{x_n\}_{n \in \mathbb{N}} \subset X$ which has a limit point $a$ but no subsequence converging to $a$. If I regard $\{x_n\}_{n \in \mathbb{N}}$ as a net over $\mathbb{N}$ directed with the standard order I know there must be a subnet converging to $a$, however I don't quite understand how this can happen.
I know there are subnets of $\{x_n\}_{n \in \mathbb{N}}$ that are not subsequences, and I have in mind the following example: $$\{x_{\left \lfloor{r}\right \rfloor}\}_{r \in \mathbb{R}}$$ This subnet has uncountably many terms but for any $j \in \mathbb{N}$ all the terms of the subnet between $x_j$ and $x_{j+1}$ are equal to $x_j$. That is, the subnet above consist of all the terms of the original sequence but there are uncountably many copies of each one before reaching the next term (of the original sequence).

Ok, that was a particular case of a subnet of a sequence (which seems to be useless if we are seeking for a subnet converging to a limit point of a sequence which has no subsequence doing that job), but I can't see how different can be any other subnet of the sequence, taking into account that the subnet can't add new elements to the image of the original sequence nor change their order. I mean, what else can the subnet do, appart from adding copies of some of the original terms (just like the subnet of the example above)? And if this is the only thing it can do, how is it possible that this thing converges but when we consider the subsequence resulting of taking away all those extra copies of the original terms, it doesn't converge?


1 Answer 1


IMHO, you have to work with some subnet constructions, to get a feel for them. A subnet is a function on index sets, as a net is a function from an index set into a space, as you probably know.

So if $f: (I, \le) \rightarrow X$ ,is a net (a sequence is just a net from $(\mathbb{N}, \le)$, where $(I, \le)$ is a directed set ($i \le i$, $i \le j \le k \rightarrow i \le k$, $\forall i_1, i_2 \in I \exists j \in I: i_1 \le j, i_2 \le j$) then a net $g:(J, \le) \rightarrow X$ is a subnet of $f$ iff there exists $h: J \rightarrow I$ that is order preserving ($j_1 \le j_2 \rightarrow h(j_1) \le h(j_2)$ and which has a cofinal image ($\forall i \in I, \exists j \in J: h(j) \ge i$) such that $f \circ h = g$.

There are other notions of subnet that are not equivalent, see this question and answer, but I find this one the easiest to understand, as it looks the most like a traditional subsequence. It's in fact exactly a subsequence if we'd demand $J = \mathbb{N}$ as well. But $J$ can be much bigger.

E.g. in your example you use the map $\mathbb{R}^+ \rightarrow \mathbb{N}$ defined by $x \rightarrow \lfloor x \rfloor$, which satisfies the requirements if both sets have their usual orders.

To get a subnet converging to $a$, we usually take a directed set that is related to $a$: let the directed set be $I =\mathscr{N}_a$ all open neighbourhoods of $a$, ordered by reverse inclusion $i_1 \le i_2$ iff $i_2 \subseteq i_1$. Directedness follows from standard inclusion properties and the fact that the intersection of two neighbourhoods of $a$ is again a neigbourhood of $a$.

Then we use that $a$ is a limit point (every neighbourhood of $a$ contains infinitely many points of the sequence, really an accumulation point) of $\{x_n: n \in \mathbb{N}\}$ to make a subnet: for each $U \in \mathscr{N}_a$ define $f(U) = \min \{n: x_n \in U\}$ and set $g(U) = x_{f(U)}$ where $f$ is well-defined as the minimum of a non-empty subset of the (well-ordered) set $\mathbb{N}$. This defines a net $g: I \rightarrow X$, and $f$ shows it's clearly a subnet of the orginal sequence in the above sense (a smaller neighbourhood can only have a larger minimal index, so order is preserved, and the image is cofinal as every neighbourhood must intersect infinitely many elements of the sequence). It also converges to $a$ as any open neighbourhood $O$ of $a$ is in the index set, so we have $x_O \in O$ and whenever $O' \ge O$ we actually have $O' \subseteq O$ so $x_{O'} \in O' \subseteq O$ as well. So every neighbourhood defines it own "tail" sets (all neighbourhoods inside it) in this net of neighbourhoods $I$; the order on $I$ is not linear, that's where you have to broaden your intuition, sets of the form $\{i \in I: i \ge i_0\}$ can be very thin threads in $I$, not "almost all points", like for sequences. So convergence wrt a net on such an order is a bit strange.

If $a$ would have a countable local base we could just take a countable base as $I$ which is decreasing and $I$ just becomes a copy of $\mathbb{N}$ and we get a subsequence instead of a subnet. To get something really different you normally use large products like $[0,1]^\mathbb{R}$ or spaces like $\beta \omega$, where the points usually have quite complicated neighbourhood structures, and many non-compatible neighbourhoods exist (for 2 neighbourhoods one need not be a subset of the other: in metric spaces neighbourhoods are essentially countable linear structures, consider the balls $B(x,\frac{1}{n})$ for fixed $x$, etc. that is why sequences suffices there.)

This question and answer also illustrate nicely how we can not a have subsequence but still a subnet that converges.


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