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Let $\tau$ be to topology on $\mathbb{R}$ with sub-basis consisting of all half open intervals $[a,b)$.

How would you find the closure of $(0,1)$ in $\tau$?

I'm trying to find the smallest closed set containing $(0,1)$ in that topology but then I realised I don't fully understand what an 'open' interval is. Is an open interval in this topology one that is half open like in the sub-basis?

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I suggest you start by finding a basis for the topology. You get a basis by taking finite intersection of elements of the subbasis. –  Daan Michiels May 15 '12 at 11:11
    
Any intersections of the sub-basis will be half open? –  user26069 May 15 '12 at 11:12
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If all else fails you may still answer the question by "Asking on math.stackexchange.". I am sure they would at least give partial credit. –  plm May 15 '12 at 11:14
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My aim isn't to answer this question per se, it's to understand more about topology by getting help with answering this question –  user26069 May 15 '12 at 11:21
    
Just to avoid misunderstanding, I was joking, I really did not mean any criticism or bad vibe. –  plm May 15 '12 at 11:50

3 Answers 3

Hints:
(i) Show that $(-\infty,b)$ is in $\tau$ for every $b$.
(ii) Show that $[a,+\infty)$ is in $\tau$ for every $a$.
(iii) Deduce that $[a,b)$ is closed in $\tau$ for every $a\lt b$.
(iv) Show that $a$ is in the closure of $(a,b)$ with respect to $\tau$ for every $a\lt b$.
(v) Conclude that the closure of $(a,b)$ is $[a,b)$ for every $a\lt b$.

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By definition of a sub-basis, a set is open iff it is a union of sets which are finite intersections of sets in the sub-basis. Here if you play around though, you can see that the intersection of two of these intervals is another such interval, or is empty.

You should look at basic open sets containing 0, here. Must they always intersect (0,1)? Draw a conclusion about 0 being in the closure or not. You should be able to spot the closure very quickly if you grasp this.

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up vote 1 down vote accepted

Recall that since the collection $S$ of all elements of the form $[a,b)$ is a sub-basis for your topology $\tau$ it follows that the collection $\mathscr{B}$ of all finite intersections of elements of $S$ is a basis for your topology $\tau$. Now it is easy to see that the finite intersection of elements of $S$ is either the empty set or another element of $S$.

Call $A = (0,1)$. Then since your topology is now given in terms of a basis, an element $x \in \Bbb{R}$ is in $\overline{A}$ iff every basis element about $x$ intersects $A$. Now it is clear from this definition that no $x < 0$ can be in the closure of $A$ because given any $x<0$, there exists $\epsilon \in \Bbb{R}$ such that $x < \epsilon < 0$ and hence $x$ is in the basis element $[x,\epsilon)$ that clearly does not intersect $A$.

We see similarly that no real number $x > 1$ can be in the closure of $A$. Now $1$ cannot be in the closure because there exists a basis element such as $[1,2)$ that contains $1$ and is completely disjoint from $A$. From these results we deduce immediately that the closure of $(0,1)$ with respect to the topology $\tau$ is the interval $[0,1)$.

Edit: You're making the mistake of assuming that being closed/open are mutually exclusive. Let us show that $[0,1)$ is closed by showing its complement is open. Now the complement of $[0,1)$ is $(-\infty,0) \cup [1, \infty)$. Now $(-\infty,0)$ is open because we can write it as

$$(-\infty,0) = \ldots \cup [-1,-0.5)\cup [-0.6,0)$$

while $[1,\infty)$ is clearly open. The union of two open sets is open from which it follows that $[0,1)$ is closed.

Now let us justify the existence of sets that are open and closed at the same time. This comes from the fact that $\Bbb{R}$ with your topology $\tau$ is disconnected. We can write $\Bbb{R}$ as $C \cup D$ where

$$C = \ldots \cup [0,1) \cup [2,3) \cup [4,5)\ldots \cup$$ and $$D = \ldots \cup [-1,0) \cup [1,2) \cup [3,4)$$

and $C,D$ are clearly open with $C\cap D = \emptyset$.

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I understand how we get to the interval $[0,1)$ that's all well and good but I thought that the closure of a set had to be closed? And as $[0,1)$ is in the basis then surely it's open? –  user26069 May 15 '12 at 11:38
    
@morphism Please see my edit above. –  user38268 May 15 '12 at 11:53

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