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Let $E$ be a topologic vector space. $x_i$ with $i\in I$ is a Cauchy net iff $\lim_j \ x_{\phi(j)}-x_{\psi(j)}=0$ for any $(\phi,\psi)$ pair of cofinal and increasing maps from $J$ to $I$.

I proved the necessity part but I struggle with sufficiency. I assumed the given limit is zero but $x_i$ is not a Cauchy net. Then $\exists V$ $0$-neighborhood $\forall i\in I \ \ \exists i_1,i_2 \geq i$ such that $x_{i_1}-x_{i_2} \notin V$. Also given limit says $\forall V$ $0$-neighborhood $\exists j_0 \in J$ such that $\forall j\geq j_0 \ \ x_{\phi(j)}-x_{\psi(j)}\in V$

I cannot see the relation between $i_1,i_2$ and maps for getting contradiction because I am not sure about surjectivity. How can I proceed to get given limit is not zero?

I appreciate any kind of help.

Thanks

Martin Sleziak
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Elias
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1 Answers1

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This is a similar question as Equivalent condition for Cauchy Sequences.

We have to prove that if the limit-condition is satisfied, then $(x_i)$ is a Cauchy net. Let us do it by contraposition.

So assume that $(x_i)$ is not a Cauchy net. This means that there exists an open neigborhood $V$ of $0$ such that for each $i \in I$ there exist $\phi(i), \psi(i) \ge i$ such that $x_{\phi(i)} - x_{\psi(i)} \notin V$. This gives us functions $\phi, \psi : I \to I$ which are cofinal and we have $x_{\phi(i)} - x_{\psi(i)} \notin V$, thus $\lim_{i \in I}(x_{\phi(i)} - x_{\psi(i)}) \ne 0$. Unfortunately there is no reason why these functions should be increasing.

What can be done to get increasing cofinal functions on some $J$?

Let $J $ be the set of all finite non-empty subsets of $I$, ordered by inclusion. This is a directed set. Note that $I$ can be canonically identified with the subset $I' \subset J$ of one-element subsets ($i \equiv \{i\}$), but the original order of $I$ is forgotten in this identification. Regarding the above $\phi, \psi$ as functions $I' \to I$, we shall extend them to functions $J \to I$. The construction will be performed by induction on the number $n$ of elements of $S \in J$.

$n=1$: If $S = \{i\}$, take $\phi(S) = \phi(i), \psi(S) = \psi(i)$.

Assume $\phi(S), \psi(S)$ have been defined for all $S$ with $\le n$ elements such that

  1. $\phi(S') \le \phi(S)$ and $\psi(S') \le \psi(S)$ if $S' \subset S$
  2. $\phi(S) \ge i$ and $\psi(S) \ge i$ for all $i \in S$
  3. $x_{\phi(S)} - x_{\psi(S)} \notin V$

To define $\phi(T), \psi(T)$ for a set $T$ with $n+1$ elements we consider the set of all subsets $S_1,\ldots, S_{n+1} \subset T $ having $n$ elements. Since $I$ is directed, we find $k \in I$ such that $k \ge i$ for all $i \in T$ and $k \ge \phi(S_r), \psi(S_r)$ for $r = 1,\ldots, n+1$. Define $\phi(T) = \phi(k), \psi(T) = \psi(k)$. This definitions assures that 1. - 3. are satisfied.

This construction yields increasing functions (condition 1.) which are also cofinal (they are even cofinal on the subset $J_1$). Clearly 3. implies $\lim_{S \in J}(x_{\phi(S)} - x_{\psi(S)}) \ne 0$.

Paul Frost
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  • First, thank you so much for your answer. I’d like yo ask you something if you don’t mind. As I understand, you constructed both J and $\phi,\psi$ pair. I thought $I$ and $J$ as fixed where $\phi$ and $\psi$ are arbitrary. If we can choose $J$, don’t we get the desired one as using your first proof? Because we can take $J:=I$ and $\phi$, $\psi$ as identity functions that are co-final and increasing. So we found a pair of cofinal and increasing functions with the given limit is not zero if $x_i$ is not a Cauchy net. – Elias Jun 14 '22 at 12:06
  • @Elias $J$ may vary. Actually one considers all possible subnets $\phi, \psi : J \to X$ of $(x_i)$ (see https://en.wikipedia.org/wiki/Subnet_(mathematics)). Taking $J = I$ would be nice, but as I said there is no reason to believe that $\phi, \psi$ are increasing. If you can find an alternative construction giving cofinal and increasing functions $\phi, \psi : I \to X$, I would like to see it. This difficulty was the reason why I introduced $J$ as the set of all finite subsets of $I$. – Paul Frost Jun 14 '22 at 12:23
  • I thought that I could take $\phi=\psi=id$ but if they are equal then limit must be equal to 0. This does not work. For your comment, I don't understand what is the set $X$? Should not we construct $\phi,\psi: J\to I$? Regarding your answer, I almost understand your construction which is probably the best one. However I cannot see why taking $\phi(S)=\phi(i)$ satisfies conditions 1-3 for n=1? – Elias Jun 14 '22 at 22:09
  • @Elias You are right, writing $X$ was an error. Take $I$ instead. For $n=1$ condition1. is trivially satisfied, 2. and 3. by the definition $\phi({i})= \phi(i), \psi({i})= \psi(i)$. Recall that the functions $\phi, \psi : I \to I$ are taken from our preliminary attempt The "new" functions $\phi, \psi : J \to I$ extend the preliminary functions. – Paul Frost Jun 14 '22 at 23:03