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Using the sequence definitions, if $X$ is compact, prove that $X$ is complete.

The definition of "compact" to be used is:

$(X,\ d)$ is compact iff every sequence in $X$ has a subsequence that converges to a point in $X$.

Also,

$(X,\ d)$ is complete iff every Cauchy sequence in $X$ converges to a point in $X$.

My attempt:

Suppose $\{x_n\}$ is a Cauchy sequence in $X$. We need to prove that $\{x_n\}$ converges to a point in $X$.

Since $X$ is compact, $\{x_n\}$ has a subsequence $\{x_{n_i}\}$ that converges to a point $x\in X$. We need to prove that $x_n\to x$ but I don't know how to proceed.

PS: Please re-open this question. I'm trying to prove this using the sequence definitions only, not using the Heine-Borel theorem.

Martin Sleziak
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Siddhartha
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    Use the Cauchy condition to prove the whole sequence converges to $x$. – Angina Seng Oct 30 '18 at 20:31
  • Let $\epsilon>0$. We need to prove that $\exists N\in\mathbb{N}$ s.t. $d(x_n,\ x)<\epsilon$ for $n\ge N$. Since ${x_n}$ is Cauchy, $\exists N_1\in\mathbb{N}$ s.t. $d(x_n,\ x_m)<\epsilon$ whenver $n,\ m\ge N_1$. How do I proceed? – Siddhartha Oct 30 '18 at 21:25
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    Suppose $x_{n_k}$ converges to $x$ as $k$ tends to infinity. Pick $N_2$ such that $k \ge N_2$ implies that $d(x_{n_k}, x) < \varepsilon$. Then there is certainly such a $x_{n_k}$ with $k \ge N_2, n_k \ge N_1$, and then $d(x_n, x) < 2\varepsilon$ for any $n \ge N_1$, using the triangle inequality via the subsequence point. Pick $\varepsilon \over 2$ at all previous stages to get exactly $\varepsilon$ here. – Henno Brandsma Oct 30 '18 at 22:04

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