If (V,||) is a normed vector space and $X\subset V$ compact, does every Cauchy sequence have a limit in X

$(V,||)$ is a normed vector space the 3 norm axioms in this case being for elements f,g from V: $$||f||\ge 0, ||f||=0 \Leftrightarrow f=0$$

$$||af||=|a|||f||, \text{for every scalar a}$$ $$||f+g||\le ||f||+||g||$$

and X is a compact subset of V, every sequence in X contains a subsequence which converges to a point in X.

Every cauchy sequence $(a_n)_{n\in \mathbb{N}}$ fulfills $$n,m\in \mathbb{N} , \epsilon>0 : ||a_n-a_m||<\epsilon$$ for a certain index $N\in \mathbb{N}$.

There is one convergent sub sequence: $a_k \rightarrow a$ with $a \in X$

Is $(a_n)_{n\in \mathbb{N}}$ Cauchy, $(a_k)$ a subsequence of $(a_n)$ then

: $$||a_k-a||< \frac{\epsilon}{2}$$ for a $k >N_1\in \mathbb{N}$

and $$||a_n-a_m||<\frac{\epsilon}{2}$$ for $m,n> N_2$

by choosing $N_z := max \{N_1,N_2 \}$, $k> N_z$:

$$||a_n-a|| \le ||a_k-a||+||a_n-a_m|| < \epsilon$$

One can say that every Cauchy sequence has a limit in X. Does this also hold if X is a arbitrary metric space?

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Yes, this is also true for arbitrary metric spaces: simply change $\|x-y\|$ by $d(x,y)$, where $d$ stands for the metric. Concisely, every compact metric space is complete.
But you don't know in advance that $V$ is complete. – Matemáticos Chibchas Jan 16 '13 at 14:33