Take the subset of the sequence space with sup norm, \ell^\infty, such that x_n belongs to [1 - 1/n,1] for all n. Does this form a compact set with respect to the norm topology?
I think this set is complete and totally bounded, so it is compact.
Yes, in this example the topology on the product coincides with the product topology (as the diameters go to $0$). And a product of compact spaces is compact.
Using a diagonal extraction, you can also prove it by hand: extract a converging subsequence of any sequence in this set.
If one shifts this set by vector (-1,-1,-1, ...) then one becomes the set A of those vectors x=(x1,x2,...) that xn belongs to [-1/n, 0] for all n. This A is in fact a subset of c0 and satisfies the criterion of compactness in this space (it is closed and is majorated in modulus by an element from c0). Since c0 is a subspace of \ell^\infty, the set is compact in \ell^\infty as well. The shift is an isometry, so it preserves compactness.
In a Banach space $X$, a set $K$ is compact if it is closed and bounded and for every $r>0$ there exists a finite dimensional subspace $F$ (which may depend upon $r$) such that $K\subseteq F + r B_X$. In this case one may take $F_n$ to be the subspace where only the first $n$ coordinates are/maybe nonzero and $K$ is the translated set described by V. Kadets above.
- Badges
- Science topic
- Similar topics
- Biological Science
- Biochemistry
- Biomolecules
- Peptides