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Let $S$ be the set of real sequences, and $S_0\subset S$ be the set of convergent real sequences. For each $f:\mathbf R\to\mathbf R$, we define $\bar{f}:S\to S$ by $$\bar{f}(\{x_n\})=\{f(x_n)\}.$$ It is known that

if $f$ is continuous, then for each $\{x_n\}\in S_0$, $\{\lim f(x_n)\}\in S_0$ and $\bar{f}(\lim x_n)=\lim f(x_n).$ ----(#)

So in some sense, $f$ and $\lim$ commute for continuous $f$.

My question is, can we reverse the process, starting with the definition of continuous functions and use this commutativity to define the notion of the limit of a sequence? In other words, is there a largest set $S_0\subset S$ (i.e. contains all other possible $S_0$) and a unique function $\lim:S_0\to\mathbf R$ such that (#) holds?

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    $\begingroup$ $S_0=${constant sequences} works also... $\endgroup$
    – abx
    Commented May 13, 2017 at 14:04
  • $\begingroup$ @abx: good catch, seems that we need to add some restriction then $\endgroup$
    – JSCB
    Commented May 13, 2017 at 14:27
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    $\begingroup$ $S_0 = S$ and the projection onto the first term of the sequence work. Obviously it is not then unique, but it never is (for any $S_0$ and $i$, projection onto the $i$th term gives a function that commutes, by your definition of $\overline{f}$) $\endgroup$
    – Maxime Ramzi
    Commented May 13, 2017 at 15:52
  • $\begingroup$ What can be added is the independence of the limit (and of the set $S$ of admissible sequences) of any finite number of elements in the sequence. Will you accept such addition? $\endgroup$
    – fedja
    Commented May 13, 2017 at 16:30
  • $\begingroup$ @ fdeja, yes, that is reasonable. $\endgroup$
    – JSCB
    Commented May 14, 2017 at 3:42

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