Proving no finite basis of the system of neighborhoods at $a$ in the real line exist.

Question

I'm not sure how to prove it, the gist is:

I need to find the "smallest" neighborhood in the basis, take a ball of half that radius and show "look, there is no member of the basis in this ball, thus we have a contradiction" However I am not sure how to say "smallest"

I want to say "choose the smallest $\delta$ of the values of $\delta$ such that $B_\delta(a)$ is maximal in each member of the basis (that is the smallest $\delta$ from a selection of $\delta$s which are the largest balls you can puff up inside each member of the basis.

Can somebody help me make this more formal, the gist is there, but it's not a proper proof (because for one I am using the words "puff up"...)

Answer 1

Each of your finitely many $U_i$ is a neighbourhood of $x$, hence by definiiton contains an open ball $B_{r_i}(x)$ for some positive number $r_i$. Pick one for each $i$. Let $r$ be the minimum of these finitely many positive real numbers, so $r$ is itself a positive real number. Consider $B_{r/2}(x)$. As $B_r(x)\setminus B_{r/2}(x)$ is nonempty, we have thus found a proper subset of all $U_i$, i.e. none of the $U_i$ is congained in $B_{r/2}(x)$. In fact, not even $\bigcap U_i$ is.

Answer 2

I want to mainly mention a point that hasn't been made in the other answers so far.

If $X$ is any topological space and $x \in X$ has a finite neighborhood basis, then it has neighborhood basis of size one.

This is because the intersection of any finite family of neighborhoods of $x$ is also a neighborhood of $x$, and is a subset of each of those begun with. So if we started out with a neighborhood basis for $x$, their intersection will by itself also form a neighborhood basis for $x$.

This is all to say that $x \in X$ has a finite neighborhood basis iff it has a unique smallest (according to $\subseteq$) (open) neighborhood.

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So it suffices to show that no point of $\mathbb{R}$ has a smallest (open) neighborhood (under the usual metric/topology).

This is basically what has been accomplished in the other answers: If $U$ is an open neighborhood of $x$, then there is an $\epsilon > 0$ such that $( x - \epsilon , x + \epsilon ) \subseteq U$. But now consider $V = ( x - \frac{\epsilon}{2} , x + \frac \epsilon 2 )$. Clearly $V$ is an open neighborhood of $x$, and $V \subseteq U$, but there are points of $U$ which are not in $V$ (for example, $x + \frac \epsilon 2$). Thus $U$ could not be the smallest open neighborhood of $x$.