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As a byproduct of a discussion with a friend, I tried to find the properties of the real line if you define neighbourhoods by closed intervals. However I quickly got stuck with this problem. So to the questions.

Can one define neighbourhoods by closed intervals? I do not see why not.

Are the intervals of the form (a,b) now considered closed?

Can [a,a]={a} be considered a neighbourhood?

Is it metrizable? My guess is no but I cannot see why it shouldn't be second countable.

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 I think this question is better suited to Math StackExchange, as it is too easy for this site. Briefly, all points are closed (e.g. $a = [b,a] \cap [a,c]$), so the topology is the discrete topology – David White Oct 26 2011 at 22:28
For a more interesting topology you may try instead the half-closed intervals $[a,b)$: see en.wikipedia.org/wiki/Sorgenfrey_line – Pietro Majer Oct 26 2011 at 22:34

closed as off topic by Ryan Budney, Bill Johnson, Andres Caicedo, Joel David Hamkins, Will Jagy Oct 26 2011 at 23:17

2 Answers

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If you allow $[a,a]$ as a neighbourhood of $a$, then the topology is discrete and there is nothing more to say (in particular it is metrizable).

If you only allow non-trivial closed intervals, then you have a finer topology, because every $]a,b[$ is still open. But every open set for this new topology is also open for the usual topology (because there must be a non trivial closed interval around every point, so there is also a non trivial open interval), so the topology you get is in fact exactly the usual topology of $\mathbb{R}$.

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The latter part of your answer is not correct, since the intersection of [a,b] with [b,c] is {b}, and so the topology is still discrete even when you insist that the closed intervals are nontrivial. – Joel David Hamkins Oct 26 2011 at 23:13
You can't enforce "nontrivial closed intervals" because of David White's comment: you can always do $[b,a] \cap [a, c]$ for $b < a < c$. – Ryan Reich Oct 26 2011 at 23:14
I assumed that the question was to generate a topology by taking the closed intervals around $a$ as a neighborhood basis at $a$. But I agree that you have also to enforce that $a$ is not one of the endpoints of the interval. – Guillaume Brunerie Oct 26 2011 at 23:28
I'm taking $\{[b,c] | b<a<c\}$ as a neighborhood basis of $a$, not an open neighborhood basis (neighborhood are not required to be open), so $[b,c]$ has no reason to be open. – Guillaume Brunerie Oct 26 2011 at 23:49
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The union of infinitely many closed intervals is not necessarily a closed interval, i.e. $\bigcup_{n = 1}^\infty [1/n, 1] = (0, 1]$, so this is not even a topology. If you do as Guillaume Brunerie and David White are implicitly assuming and generate a topology via closed intervals, you get the discrete topology. If you take the above computation as a hint and look at right-half-closed intervals or left-half-closed intervals, you get what Pietro Majer said, the Sorgenfrey topology.

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