Given a complete normed space $X=(X,\|\cdot\|)$. Every Cauchy sequence converges in it. I am not able to understand why we can't show that every bounded sequence in $X$ will have a convergent subsequence.
Please give an example to clarify why completeness does not imply compactness and do explain where does the problem lie.
It is not that every bounded sequence that needs to have a convergent subsequence, but rather every sequence in order to be sequentially compact.
Even worse, consider the unit ball in any infinite dimensional complete vector space. This is closed and bounded in a complete normed space, but not compact.
Edit: In my mind, to say that a metric space is complete recovers our intuition of the continuum. In particular, it tells us that there are no "gaps" in our vector space. For example, $\mathbb Q$ is a perfectly good vector space over itself. On the other hand, it is not so nice topologically.
compactness is some way of characterizing how "bunched" together something is. This need not agree with our usual notion of "distance." For example, the unit ball is metrically "close" (and is indeed closed, so we cannot "converge out" of it)-- yet, because we are in infinite dimensions, there is not necessarily a finite way to get from one vector to another once we've specified a covering. Hence, we can think of "compactness" as some kind of finiteness condition, even when the set in question is not itself finite.
There are much better answers here, but they are more in the realm of general topology.
$\mathbb{R}$ is complete but not compact. The problem is that the definition of compactness is "every sequence has a convergent subsequence" and not "every bounded sequence has a convergent subsequence".
Your question seems to be about the local compactness of Banach spaces.
Look at a sequence of sequences with general term $A_n=(\delta_{i,n})_{i>0}$ which is in $l^p$. What can you say about the norm of its term and the norm of the "potential" limit? By the way there is compactness in a weak sense, but you might need Hilbert spaces for their scalar product. See Banach-Alaoglu's theorem.
It is also worth mentioning that a general result exists: A normed vector space (real or complex)'s closed unit ball is compact (for the norm topology) if and only if it is of finite dimension (Riesz lemma).
A normed space $X=(X,\|\cdot\|)$ is a metric space with the metric induced by the norm \begin{align*} d(x,y)=\|x-y\| \end{align*} All concepts defined for metric spaces are applicable, in particular, to normed spaces. Therefore we can also use the results of completeness and compactness derived in metric spaces.
The following facts stated without proof might help to better see the relationship of completeness and compactness in metric spaces and therefore also in normed spaces.
The following is valid:
A metric space $(X,d)$ is complete if every Cauchy sequence in $X$ has a convergent subsequence.
In a metric space $(X,d)$ compactness and sequential compactness are equivalent.
We see that in a metric space $(X,d)$ compactness is a stronger property than completeness, since we have
\begin{align*} \text{ compact } \quad\Longleftrightarrow\quad \text{ sequentially compact } \quad\Longrightarrow\quad \text{ complete } \end{align*}
The implication in the other direction does not hold, since e.g. $\mathbb{R}$ is complete in the usual euclidean metric, but not compact.
So, it is reasonable to ask what extra condition we could impose to a complete metric space in order to guarantee its compactness. Such a condition is total boundedness.
A metric space $(X,d)$ is said to be totally bounded if for every $\varepsilon>0$, there is a finite covering of $X$ by $\varepsilon$-balls.
- Example: Under the metric $d(x,y)=|x-y|$, the real line $\mathbb{R}$ is complete but not totally bounded, while the subspace $(-1,1)$ is totally bounded but not complete. The subspace $[-1,1]$ is both complete and totally bounded.
Theorem: A metric space $(X,d)$ is compact if and only if it is complete and totally bounded.
\begin{align*} \text{ compact } \quad\Longleftrightarrow\quad \text{ complete } +\text{ totally bounded } \end{align*}
If $X$ is a compact metric space, then $X$ is complete as noted above. The fact that $X$ is totally bounded is a consequence of the fact that the covering of $X$ by all open $\varepsilon$-balls must contain a finite subcovering.
The other direction can be shown, by proving that completeness and total boundedness implies sequential compactness, from which compactness follows.
Note: The answer above is based upon §45 Compactness in Metric Spaces in Topology by J.R. Munkres.
