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Problem:

Let $ \mathbb{R}_l $ denote the topological space whose underlying set is the real line $ \mathbb{R} $ and the topology is generated by the half closed intervals $ [a,b) $. Prove that the topological space $ \mathbb{R}_l $ is totally disconnected.

My proof:

A space is totally disconnected if its only connected components are one-point sets. Given any set $ I\in \mathbb{R}_l $, $ I=[a,b) $ for some $ a\le b $ in $ R $. If $ a=b $, $ I $ is a one-point set and clearly a connected component (there are no $ A,B\in I $ both non-empty and proper). If $ a<b $, $ I $ is not a one-point set and there exists a $ c $ with $ a<c<b $. Then $ A=[a,c) $ and $ B=[c,b) $ are both non-empty, proper open subsets of $ I $ and they constitute a separation of $ I $ because $ A\cap B=\emptyset $ and $ A\cup B=I $. It is therefore clear that the only connected components in $ \mathbb{R}_l $ are the one-point sets, and hence the topological space $ \mathbb{R}_l $ is totally disconnected.

Question: Is my proof correct?

Barbara
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3 Answers3

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Your proof is almost complete, and what you have shown is done very nicely.

However:

What about sets that are not half-closed intervals?

Your proof only shows that all intervals are not connected components. It should be easy to complete it by showing that

  • closed intervals are not connected
  • open intervals are not connected
  • if a set $A\subseteq \mathbb R$ is not an interval, it also cannot be connected.

Hint:

$X\subseteq$ is not an interval if and only if there exist some $a<b<c$ such that $a\in X$, $b\notin X$ and $c\in X$

5xum
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  • Thank you, I see now that my proof is not complete. I must also show that if a set is not an intervall, it cannot be connected. But why must I show open and closed intervals as well? Only half-closed intervals (in this case closed to the right and open to the left) and unions of these are in the topology. – Barbara Nov 23 '16 at 12:23
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    @GjermundJohansen You have to show that the only connected component of the set are singletons. Thus, you must show for every other set that it is not a connected component. Not only open sets, but for all sets. For example, your proof must also show that $\mathbb N$ is not a connected component. So, if you take an arbitrary set, it can either be an interval or not, and if it is an interval, it can be open, closed, or semiclosed. – 5xum Nov 23 '16 at 12:26
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    Don't bother with intervals, you just want to show that any subset with more than one point is not connected. If A is such a set, it contains points a and b with a < b. Choose c between a and b. The intersections of A with (-oo, c) and [c, oo) are non-empty and clopen in A and separate it. Therefore it is not connected. (It doesn't matter if c is in A or not.) – David Hartley Nov 23 '16 at 14:13
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Let $S$ be a subspace of $\mathbb R_l$ with $a,b \in S$ and $a<b.$

$A=(-\infty,b)=\cup_{x<b}[x,b)$ is open in $\mathbb R_l,$ and $B=[b,\infty)=\cup_{y>b}[b,y)$ is open in $\mathbb R_l$. So $A\cap S$ is not empty (because it contains $a$) and is not equal to $S$ (because it does not contain $b$), but it is open in the subspace $S$, and it is closed in $S$ (because its complement in $S,$ which is $B\cap S,$ is open in $S.$).

So any subspace $S$ of $\mathbb R_l$ with more than one point is a disconnected space.

DanielWainfleet
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Let $F\subset \mathbb R_l$ be non empty, finite with atleast two elements. Take any $a$ in $F$. Set $d= \min_{x\in F-\{a\}} (|x-a|)$. $\{a\}= F\cap [a,a+d/2)$. So $\{a\}$ is open in subspace topology. $F$ is a discrete space so $F$ can't be connected.

If $S\subset \mathbb R_l$ is infinite, then take any $s$ in $S$ such that $s> \inf S$. $S= ((-\infty, s)\cap S)\cup ([s,\infty)\cap S)$ is a separation so not connected.

So the only connected subsets of $\mathbb R_l$ are the singletons hence $\mathbb R_l$ is totally disconnected.

Koro
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