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Consider the quotient space $X=mathbbR^2/(mathbbR times 0)$ I'm supposed to prove it isn't first countable. It seems to me that this space is homeomorphic to two open triangles joined at their apexes with that point added. That point is the problematic one, but why isn't the collection of open balls with descending rational radii centered at this point intersected with $X$ a countable basis?

Thank you for your answers.

Edit: I found this problem on my university's website. We did this same problem at school but first-countability was replaced with local compactness. It could be possible that the space is in fact first-countable, and the problem statement on the website was incorrect.

$mathbbR times 0$ is collapsed to a point $ast in X=mathbbR^2/(mathbbR times 0)$. This point does not have a countable base of open neighborhoods.

$X$ is endowed with the quotient topology. This means that open neigborhoods of $ast$ are in a $1$-$1$-correspondence with open subsets of $mathbbR^2$ containing $mathbbR times 0$. Let us denote the collection of these sets by $mathfrakU$.

Therefore, as YuiTo Cheng remarked in his comments, we have to show that there does not exist a countable collection of sets $U_ n in mathfrakU$ such that for every $U in mathfrakU$, $U_n subset U$ for some $n$.

Let us consider any countable collection of $U_ n in mathfrakU$. For each $n in mathbbN$ there exists a number $u_n > 0$ such that $(n,u_n) in U_n$. The set $D = (n,u_n) mid n in mathbbN $ is closed in $mathbbR^2$, thus $U = mathbbR^2 setminus D in mathfrakU$. But by construction no $U_n$ can be contained in $U$.

Remark:

This generalizes as follows. Let $X,Y$ be topological spaces such that $X$ contains an infinite subset $A subset X$ without limit points and let $y in Y$ be a point such that $ y $ is not open in $Y$. Then $(X times Y)/(X times y )$ is not first countable.