Topology proof

>since it's a metric space

so the two point-space with metric d(p1, p2)=1 is also connected?

>all metric spaces are connected
Wew
Consider the metric space ({0,1},d) where d(0,0)=d(1,1)=0 and d(0,1)=d(1,0)=1. Then taking as open those sets containing x such that d(x,a)

Lmao, beat me to it.

>Consider w(t) = a*(1-t)+t*b for t in [0,1].
If we're considering a vector space over Q, tb and (1-t)a are not necessarily well-defined elements of the space, no? Say we let t=sqrt(2)/2
However every vector space over a connected field is connected, say, real or complex vector spaces. Which makes sense, as you can "stretch" or "shrink" vectors over arbitrary "connected" ranges. So I think this proof works in those cases.

Makes sense. Maybe OP means a Banach space? So that it is complete.

What I mean is the proof works for the fields R and C, since tb and (1-t)a and thus their sum are clearly elements of the space by hypothesis. I don't know if the method generalizes since I've never done anything with """topological fields,""" but wikipedia says the result holds, so

A Banach space is already over R or C by definition so the result should hold, and for Hilbert spaces as well, since they're also Banach spaces. How about arbitrary inner product spaces?
Since OP can't be arsed to ask his question properly

Completeness is not necessary, but being over R or C is.

>being over R or C is
Are there any nontrivial connected fields other than R and C, though? And isn't being over R or C a sufficient condition, or is there a counterexample to one or both?

>Are there any nontrivial connected fields other than R and C, though?

I don't think so.

>And isn't being over R or C a sufficient condition

For connectedness, yes.

I'm doing it for a topological space 'cause easy.
From the definition of a topology T on a set X:
T contains X and the empty set.

Any set in the topology on X is open by the definition of open sets. The complement of an open set is a closed set. Therefore X and the empty set, being complements of eachother, are both open and closed.

Please sweet jesus, give me the strength to resist this bait

First is a quantifier mistake. The space X is connected if there are no proper subsets which are both closed and open. You can't just point out that X and 0 are open and closed in every space by the definition of topology, because that alone isn't sufficient to prove connectedness. You need to show there are no others. That every simultaneously open and closed subset of X is either X or 0. Yeah, every space is connected if you use the trivial topology (X and 0) since those two are the only open sets, period. But that's the coarsest structure possible and in a certain sense is just a one-point space.

Yes because it's a metric space and by theorem 8829996 it is connected.

what? they are not. Considere the set of integers, any subset is both open and closed for the classical metric.

>theorem 8829996
Lost hard

Here you have just shown that given an arbitrary topology over an arbitrary set it must be the case that both X and the empty set are both close and open.

You have however NOT proven that they are the only sets that are both closed and open. In fact, this is not true in general and many people have pointed out counter examples. Given an arbitrary set it is also possible to trivially construct a topology where the statement fails (eg. Given the set X, let the power set of X be a topology over X, then every subset of X is both closed and open).

Your assessment isn't wrong but it's totally not even close to the misunderstanding in that user's post.

>Here you have just shown that given an arbitrary topology over an arbitrary set it must be the case that both X and the empty set are both close and open.
>You have however NOT proven that they are the only sets that are both closed and open.
That's what I meant by "quantifier mistake."