A subset K ⊂ X of a metric space is covering compact if for each open cover {Uα}α∈A, there is a finite subset of opens from the cover, {Uα1,...,Uαn}, such that K ⊂ Uα1 ∪···âˆªUαn. In other words, every open cover has a finite subcover
Something is sequentially compact if every sequence has a convergent subsequence. You can define seq. compness for general topological spaces. You can define (covering) compact for general topological spaces. It so happens the definitions are equivalent for metric spaces.
More generally, they are equivalent for second countable spaces.
@PedroTamaroff (for whenever you get back) you have "kets" which are vectors written $\mid a \rangle$, where $a$ is whatever you name the vector. "bras" are conjugate transposes of vectors, written $\langle b \mid$. when you multiply a bra and a ket, you get $\langle b \mid \mid a \rangle = \langle b \mid a \rangle$ i.e. the inner product of $b$ with $a$
We have two tags quadratics and quadratic-equation. Should not they be synonyms? (I do not have sufficient reputation in either of the two for suggesting a synonym. But as two mods are present in the room, I thought I might bring it up here.)
The tag-wiki of the first one specifically says that it is for questions about quadratic equations. The second one has empty tag-wiki, but the name of the tag is self-explanatory.
Anyway, knowing that someone reads old threads is a good thing. It is nice to know that the answer I wrote can be useful for someone. However, just checking the number of views on some old questions would give me the same information, I do not need a downvote to know that "someone was there".
@SabyasachiMukherjee I think that this is not exactly a conclusion you can draw, since now we're including you. As you learn the machinery, you will deepen your understanding, which will make solving those problems easier.
@Hawk that Q on main about uniqueness of representation of p as sum of two squares .. can you show that for non unique representation case .. p becomes composite ?
@Hawk take for example if $x=gcd(a,c)$ and $y=gcd(b,d)$, then $2(a^2+b^2+c^2+d^2) = (x^2+y^2)(x'^2+y'^2)$, for some $x',y'$ which I leave you to figure out :)
@Hawk the factorization is easy .. standard corollary that appears alongside with Fermat's proof of representation of primes of form $4k+1$ as sum of two squares with infinite descent :)
