I have a quick question regarding the "$\Leftarrow$" part of the proof, i'm not sure why something is true. It's a proof my professor write, and i'm trying to understand it. I'll just write out the proof until i get to my problem.
Assume if any vector sequence $ \in F$ converges, it's limit is also $\in F$.
Now i have to show $F$ is closed, or $F^c$ is open.
Let's have a look at any element in in $F^c $, call it $x$ so $x \in F^c$.
I have to find an $r>0$ such that $B(r,x) \subset F$
Let's assume $\nexists r$.
$\Rightarrow \exists B(r,x) \not\subset F^c$
- $\Rightarrow \exists B(r,x) \cap F \not= Ø$
- Since it's not empy, pick elements from there and call it $x^{(k)}$
- We now have a vector sequence $ \{x^{(k)} \}^{\infty}_{k=1}$ where $x^{(k)} \in F \forall k \in \mathbb{N}$
And here is my problem. It supposbly should be true, that: $0 \leq x^{(k)} \in F \forall k $
Does anyone know why this inequality holds?
The first 5 bullets are okay.
In fifth bullet is is assumed that so such $r>0$ exists.
That means that for every $r>0$ the set $B(r,x)\cap F$ is not empty.
Now e.g. for every $k>0$ pick some $x^{(k)}\in B(\frac1k,x)\cap F$.
Then $x^{(k)}\to x$ so that we may conclude that $x\in F$.
Now a contradiction has been reached and the conclusion is that the assumption that no such $r>0$ exists is wrong.
Then you are ready with proving that $F^{\complement}$ is open.
The inequality $0\leq x^{(k)}$ seems to "fall from the sky" and makes no sense at all in this context.