Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$(V,||)$ 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?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
Thank you very much. – bakabakabaka Jan 16 '13 at 14:57

A compact subset of a metric space is closed. Closed subsets of complete spaces are complete. By definition of completeness, every Cauchy sequence in that space has a limit in that space.

share|cite|improve this answer
But you don't know in advance that $V$ is complete. – Matemáticos Chibchas Jan 16 '13 at 14:33
Apologies, I misread the question as assuming completeness. – anonymous Jan 16 '13 at 19:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.