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If $f:X\to Y$ takes Cauchy sequence to Cauchy sequence then prove that $f$ is a continuous function.

Let $x_n$ be a sequence in $X$ such that $x_n\to x\implies x_n$ is Cauchy $\implies f(x_n)$ is Cauchy but that does not guarantee that $f(x_n) \to f(x)$ .

So how is the above result true. Please help.

Martin Sleziak
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If $(x_n) \rightarrow x$, make a new sequence $y_{2n} = x_n, y_{2n+1} = x$, so intersperse terms of the sequence with the limit.

  1. Show that $(y_n)$ is Cauchy.
  2. So the sequence $(f(y_n)) = f(x_0), f(x), f(x_1), f(x),\ldots$ is Cauchy by assumption.
  3. From this show that $f(x_n) \rightarrow f(x)$.

For the last, there is a more general fact you might know: if a Cauchy sequence has a convergent subsequence (with limit $p$), the whole sequence converges to $p$ as well. But a direct proof is also easy enough.

Henno Brandsma
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    This is hands down elegant and brilliant explanation! Thanks a lot. – Error 404 Dec 02 '17 at 06:42
  • @Mathmore glad you like it! – Henno Brandsma Dec 02 '17 at 07:23
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    Beautiful proof! Could the same steps be used to show that a function that takes convergent sequences to convergent sequences is continuous? Or that a function that takes convergent sequences to Cauchy sequences is continuous? – Anu Feb 17 '18 at 11:15