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This question involves finding a base for the neighbourhood system at $0$, as well as at each point other than $0$ in the either-or topology.

The either-or topology is defined on the set $X = [-1,1]$ as follows (just to refresh):

$U$ is open $\Longleftrightarrow 0\notin U$ or $(-1,1)\subseteq U$.

Here goes my thought process...

The nbd system at $0$ would be the collection of all nbd's of $0$ and these would be intervals in $X$ containing open sets containing $0$. Seeing as these open sets must contain $0$, they must be $U$ sets of the form

$(-1,1)\subseteq U$.

So would my saying that the nbd system of $0$ would simply be $X$? What about $(-1,1)$ - trivially I am assuming that holds?

As for finding a base for this neighbourhood system I am completely lost. I am not satisfied by my level of understanding of bases of nbd systems so any suggestions as to good, comprehensible literature on that topic would be more than welcome.

Thank you in advance!

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2 Answers 2

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You’re right that the open nbhds of $0$ are the $U\subseteq X$ such that $(-1,1)\subseteq U$, but you missed some of them. There are exactly four such sets: $(-1,1),[-1,1),(-1,1]$, and $[-1,1]=X$ itself. This is the entire nbhd system at $0$, and you notice that every one of these sets contains $(-1,1)$. In other words, $\{(-1,1)\}$ is a base at $0$: for any open set $U$ containing $0$, it’s true that $0\in(-1,1)\subseteq U$.

Now suppose that $x\in[-1,1]\setminus\{0\}$. Can you find a single open set that is contained in every open nbhd of $x$? If so, you can get a one-element nbhd base at $x$, too.

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  • $\begingroup$ @katherinebarry: That wouldn’t give you one set, though: you’d need to look at more than one $\epsilon$. Ask your self this: if $x\ne 0$, is $\{x\}$ an open set? $\endgroup$ Commented Mar 9, 2013 at 19:30
  • $\begingroup$ @brianmscott no it wouldn't be an open set $\endgroup$ Commented Mar 9, 2013 at 19:34
  • $\begingroup$ the only singleton open set would be {0} surely? $\endgroup$ Commented Mar 9, 2013 at 19:35
  • $\begingroup$ @katherinebarry: Just backwards. $0\notin\{x\}$, so be definition $\{x\}$ is open. And we already saw that the only open sets containing $0$ are the four that I mentioned in my answer, so $\{0\}$ is not open. $\endgroup$ Commented Mar 9, 2013 at 19:37
  • $\begingroup$ @brianmscott oh of course, how stupid of me i'm not sure what i was thinking! Thanks for your help, hopefully I can figure it out from here! $\endgroup$ Commented Mar 10, 2013 at 9:12
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Hints: First note that open sets in the either-or topology need not be intervals (or even disjoint unions of intervals). For example, the set $\{ -1 \}$ is open, as well as $( -1 , \frac{-1}{2} ] \cup ( 0 , \frac{1}{2} ]$. So the first thing to do is to (sort of) forget about open intervals. That said, if $U$ is an open neighbourhood of $0$, then clearly $0 \notin U$ is false and the conditions on the topology leave very few possibilities. Go through these to determine which are actually open. You might even be able to find a smallest open neighbourhood of $0$ (and I think you have).

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