I was trying to use the language of nets and subnets. But I noticed I was essentially proving every subsequence $x_1, x_2, \ldots $ has a convergent subsequence - namely $x_1, x_1, x_1, \ldots$ - which of course isn’t allowed as a subsequence. But when couched in the language of subnets I’m not seeing where exactly my error is:

Let $\{ x_a\}_{a \in A}$ be a net, i.e. simply points of $X$ indexed by a directed set $A$. Fix some element $a_0 \in A$, and let $x=x_{a_0}$. I want to show (the obviously false statement) that there is a subnet having $x$ as a limit. So to recall the definitions: I want to show there is a directed set $B$, a monotone (co)final function $h:B \to A$ so that for the resulting subnet $\{x_{h(b)}\}_{b \in B}$, for every open neighborhood $U$ of $x$, there exists $b_0 \in B$ such that $x_{h(b)} \in U$ for all $b \geq b_0$.

Let $B$ be the set of pairs $(U, a)$ where $U$ is an open neighborhood of $x$ and $a \in A$ is such that $x_a \in U$.

We can order $B$ by declaring $(U', a')\geq (U,a)$ if $U' \subseteq U$ and $a' \geq a$. Let $h:B \to A$ send $(U,a)$ to $a$; $h$ is clearly monotone, and given $a\in A$, $(X, a) \in B$ maps to $a$ so $h$ is cofinal.

Let $U$ be a given neighborhood of $x$. Set $b_0=(U, a_0)$. Now let $b=(U', a') \in B$ satisfy $b \geq b_0$, so $x_{a'} \in U' \subseteq U$. I.e. for all $b \geq b_0$, $x_{h(b)} \in U$ as we wanted to show.

  • $\begingroup$ Where does your definition of subnet come from? There are various non-equivalent notions of subnet out there. $\endgroup$ – Michael Greinecker Aug 16 '18 at 20:55
  • $\begingroup$ @MichaelGreinecker from the Wikipedia page $\endgroup$ – usr0192 Aug 16 '18 at 20:56

As Michael Greinecker pointed out, B is not necessarily directed. It can fail to have the upper bound property.

If it does have that property, then for any open nbhd $U$ of $x$ and any $a \in A$ there will be an upper bound for $(U,a_0)$ and $(X,a)$, say $(V,b)$. So we have $b \geq a$ such that $x_b \in U$. That shows that $x$ is a cluster point of the net.

Conversely, if $x$ is a cluster point then, given $(U,a)$ and $(V, b)$ in B, first choose $c \in A$ greater than a and b and then choose $d \geq c$ such that $x_d \in U \cap V$. Then $(U \cap V, d)$ is an upper bound for $(U,a)$ and $(V,b)$ in B.

Thus B is directed iff $x$ is a cluster point of the net.


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