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Let $X_1$, $X_2$ be metric spaces and $f\colon X_1\to X_2$ be continuous. Let $(x_i)_i$ be Cauchy in $X_1$.

I argue that $(f(x_i))_i$ is Cauchy in $X_2$:

Let $\epsilon > 0$. Then there exists an $N$ such that for all $i, j\ge N$, we have $d(x_i, x_j) < \epsilon / 2$. Now take a $\delta > 0$ such that $d(y, x_N) < \delta\implies d(f(y), f(x_N)) < \epsilon / 2$. Choose a $K\ge N$ such that for all $i, j\ge K$, we have $d(x_i, x_j) < \delta$.

Now, let $i, j\ge K$. Then $d(f(x_i), f(x_j))\le d(f(x_i), f(x_K)) + d(f(x_j), f(x_K))$. Since $d(x_i, x_K), d(x_j, x_K) < \delta$, we have that $d(f(x_i), f(x_j)) < \epsilon/2 + \epsilon/2 = \epsilon$.

Question: I am aware that this is in general false, as discussed here. So, where did I go wrong?

Atom
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    The answer here is probably of interest. – Andrew D. Hwang Feb 11 '22 at 20:57
  • How do you know that such universal $\delta$ exists? (It doesn't exist, atleast not for all continuous function.) – Laci Feb 11 '22 at 20:58
  • @Laci That $\delta$ if for $x_N$. – Atom Feb 11 '22 at 20:59
  • I suppose you are trying to say that if $d(x,y)< \delta$, then $d(f(x),f(y))< \epsilon/2$, but there is no universal $\delta$ that works for every $y$. What we know is that for all $\epsilon$ and $y$ there exists a $\delta_{y,\epsilon}$ such that for all $x$ with $d(x,y)< \delta_{y,\epsilon}$ implies that $d(f(x),f(y))< \epsilon/2$. Notice that $\delta$ depends on $y$. (Sorry if my english is wonky, not my native language. :) ) – Laci Feb 11 '22 at 21:04

1 Answers1

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You have $d(y, x_N) < \delta \implies d(f(y), f(x_N)) < \epsilon/2$, but that doesn't say $d(f(y), f(x_K)) < \epsilon/2$.

Robert Israel
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