"If you model some phenomenon with a polynomial, it's often of interest to determine when the polynomial evaluates to zero. One of the tools used in deciding when this happens is factoring."
The challenge on this site to write correct answers in a way that even monkeys will understand to avoid downvotes, else write them so complicated they think they must be correct
@Matt Hah, thanks. I guess. It's an old answer... I forgot about it, too. Either way, my eyelids are dropping like frogs from the sky, and I'm not even in Egypt during the biblical period!
Had a long session with yunone on extension of uniformly continuous functions in chat, enjoyed that it's still dry inside my house despite the thunderstorms and am freezing my feet off at the moment... Great day!
Define a valid path to be a sequence $p : \mathbb N \to \mathbb N^2$ such that (i) every point in $\mathbb N^2$ appears exactly once. (ii) The sequence is non-decreasing in the sense that for all $i < j$, either $p_i$ and $p_j$ are incomparable or $p_i < p_j$.
Of course, the order on $\mathbb N^2$ is given by $(a,b) \leqslant (c,d)$ if $a \leqslant b$ and $c \leqslant d$.
@robjohn :2907989 Now, given a doubly-indexed sequence $a = (a_{mn})$ and a "valid path" p, we can associate a sum $\sum_{i=1}^\infty a_{p(i)}$ provided this series converges.
@Tim I didn't mean to say that you should keep quiet. I was referring to our two exchanges in the comments today. The last comment came almost immediately after I posted it.
@tb Sure, that topology is bizarre. But (a) what do I do, however much I dislike it, that is the coarsest topology that makes the norm continuous. (b) I said that the topology is uninteresting in my comment. =)
@tb I see now. Your just deleted comment was supposed to address Sri's comment to Bill. No, I wasn't offended. I was making joke about keeping any secret that you might imply.
@tb After reading some Wiki, I was wondering if the topology induced by a norm is the coarsest one that can make addition and scalar multiplication continuous?
