I'm having a little trouble understanding the definition of open and closed set in a metric space.
I'm going to use an excersize:
Lets consider $Y:=(-1,1]$ with the usual metric of $\Bbb R$. Decide whether these sets are open or closed in $Y$:
(ii) $(-\frac 12,0]$
(i) I say is open, since there's a proposition somewhere that says:
Every metric space is open in itself
Maybe the proof needs a little clarification for me. Says that we can take any $x\in X$ and for every $\varepsilon >0$ we get that $B_X(x,\varepsilon)\subset X$, hence every elemente in $X$ is an interior point. I don't see why this is true, if we suppouse that $X$ has an element such that it is a contact point (I belive we can assume that since the only hypothesis is that $X$ is a metric space), then we can't find any open ball that stays completly in $X$, I know I'm missing something, but I can't see what, (I know is not the same to say open in $\Bbb R$). Also, is there an analogue version of this but with closed? like, every set is closed in itself?
(ii)I say this isn't nor open nor closed, since $0$ is a contact point and $-\frac 12$ is too a contact point but is not included in the set. Am I using the definition of contact point well for these points? Also, in wich case this set would be closed? maybe if we had $[-\frac 12,0]$?
(iii)I think is closed, since it contains all of its contact points. Is it ok for me to think that $-\frac 12$ is a contact point? I think it is, since for every $\varepsilon>0$ we have that $B_X(-\frac 12,\varepsilon) \cap (-1,0]\neq \varnothing$.