Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I realise this is a very easy question. But it seems to me that from the standard (open sets) definition of a topology ($X$ and $\varnothing$ open, closed under arbitrary unions and finite intersections) that the collection $\{\Bbb R, [0, 1], \varnothing\}$ forms a topology on $\mathbb R$. Why is this not the case?

Also, can someone point me in the direction of a good proof for why an open set in a topology does not contain it's boundary points (a proof from the axioms of a topology not from the concept of an open set in a metric space).

share|improve this question
1  
What makes you think it's not a topology? –  Jason DeVito Sep 23 '12 at 1:25
    
After your most recent revision of the question, what you have is a topology. –  Michael Hardy Sep 23 '12 at 1:30
    
That's what I thought, I think I was just thrown by the examples of topologies on R induced by metrics. –  user42340 Sep 23 '12 at 3:09
    
Thanks to everyone for the help –  user42340 Sep 23 '12 at 3:09
add comment

1 Answer 1

$\{\mathbb R,[0,1],\varnothing\}$ is a topology on $\mathbb R$. Also, an open set may contain its boundary; for example, $\mathbb R$ itself. This uses the definition that the boundary of a set is the intersection of its closure and the closure of its complement, so that the boundary of $\mathbb R$ is $\varnothing$.

share|improve this answer
1  
This, of course, depends on the definition of "boundary point." The one I've seen is that $p$ is in the boundary of a set $U$ if every open set containing $p$ contains both points of $U$ and points not in $U$. By this definition, an open set can not contain any boundary points, for if $p\in U$, with $U$ open, then $U$ itself witnesses the fact that $p$ is not in the boundary. –  Jason DeVito Sep 23 '12 at 1:34
    
(Your definition doesn't avoid the issue I mentioned. If $U$ is open and $p\in U$, then $U^c$ is closed, so is its own closure, and $p\notin U^c$, so $p$ is not a boundary point. Also, perhaps I should say I'm worrying about the case when the boundary is nonempty, for if its empty, $U$ automatically contains all those points!) –  Jason DeVito Sep 23 '12 at 1:39
    
Fair enough, I'll delete all these comments in a minute. You may want to include a note about how this is the only way an open set can contain its boundary: If the boundary is nonempty, an open set never contains any of the boundary points. –  Jason DeVito Sep 23 '12 at 1:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.