Cardinality of a set of disjoint open sub intervals of $( 0 ,1)$

Question

Let $A$ be any collection of disjoint open subintervals of $(0 ,1)$ . Then what is maximum cardinality of $A$ ?

I know one easy way to prove its countable is that every open interval has rational points and if there are uncountably many disjoint sub intervals in $A$ , then $(0 , 1)$ has least uncountably many rational points , a contradiction .

I have two questions now

$(1)$ The above fact I am trying to prove in some other way around (if possible)

Let $ Λ $ be indexing set for $A$ i.e $A=\{ (a_{\alpha} , b_{\alpha}) \subset (0,1) | \alpha \in Λ \} $

Then the set $\{ a_{\alpha } , \alpha \in Λ \} $ is a discrete set (?)

Let $x_1=\inf \{ a_{\alpha} | \alpha \in Λ \} $ So $x_1=a_m $ for some $m \in Λ$

Again $a_{\alpha} > b_m >a_m \forall \alpha \in Λ\setminus m $

Thus let $x_2=\inf \{ a_{\alpha} | \alpha \in Λ \setminus m \} $

In this way ,all left end points of the subintervals can be put into one one correspondence with $ℕ$ , so $A$ must be at most countable ..

Can you help me fill up the minute details here which I am obviously missing ?

$(2)$ After a little analysing , I have the feeling that $A$ can be at most finite and not countably infinite .

Because the set $\{ a_{\alpha } , \alpha \in Λ \} $ is a bounded infinite subset of $ℝ$ and so must have a limit point in $(0 ,1)$ Then we can't get a family of disjoint open sub intervals because the end points are getting arbitrarily close.

Is this true what I am thinking ?

Please help me formalise or write a proper full proof or even show counter examples , if possible . Thanks

Answer 1

For the countably infinite case, consider defining $$ x_n := \frac{1}{2^n}, \qquad n=0,1,2,\cdots $$ to be the "edges" of the open intervals, and then $$ m_n := \frac{x_n + x_{n+1}}{2}, \qquad n = 1,2,3,\cdots $$ to be their midpoints. Notice that $(x_{n+1},x_n)$ has width $$ x_n-x_{n+1} = \frac{1}{2^n} - \frac{1}{2^{n+1}} = \frac{1}{2^{n+1}} $$ So we will choose our radius as half that, $$ r_n := \frac{1}{2^{n+2}} $$ and then $$ \left\{ \left( m_n - r_n, m_n + r_n \right) \right\}_{n=1}^\infty $$ is a collection of countably many intervals, diskoint and within $(0,1)$.

You can see it more visually in this Desmos demo. Below, I've highlighted alternating intervals per the parity of their $n$:

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To see that you can't have uncountably many, consider:

If $\{S_i\}_{i \in I}$ are such that

• $S_i \subseteq (0,1)$ for each $i$
• the $S_i$ are disjoint
• the $S_i$ are intervals

then the following are true:

• each $S_i$ has a positive length, i.e. $\operatorname{length}(S_i) > 0$ for each $i$
• $\sum \operatorname{length}(S_i) \le \operatorname{length}((0,1))=1$

Can the latter ever be achieved for an uncountable set?

Answer 2

The set of (left) endpoints can have a limit point in the closed interval $[0,1]$ that is not part of, or an end point of, any of the individual open subintervals. A simple example is the set of subintervals $(\frac1{n+1},\frac1n)$ for $n=1,2,\ldots$