Partitioning $[0,1]$ into pairwise disjoint nondegenerate closed intervals

Question

My friend threw me a challenge. He told me that he proved the follow proposition:

The topological space $[0,1]$ cannot be partitioned into pairwise disjoint nondegenerate closed intervals (except the trivial partition $\{ [0,1] \}$).

I think that it's false because I prove the follow proposition:

It's impossible to cover $[0,1]$ with pairwise disjoint closed intervals with a finite number of singletons (except for $\{ [0,1] \}$).

Proof?

Let be $0<a_1<a_2<1$, I have that $$[0,1]=[0,a_1] \cup [a_2,1] \cup \left]a_1,a_2\right[.$$

Now I have one "hole" to cover with closed interval.

$\left]a_1,a_2\right[=[a_1+r_1,a_2-r_2] \cup \left]a_1,a_1+r_1\right[ \cup \left]a_2-r_2,a_2\right[$ with $r_1,r_2>0$ such that these are disjoint.

Now I have two "holes" to cover $\left]a_1,a_3\right[$ and $\left]a_4,a_2\right[$ where $a_3=a_1+r_1$ and $a_4=a_2-r_2$.

Following in this way I obtain that $$[0,1]= \bigcup_{i=1}^{\infty}[a_i,a_{i+1}] \cup Z$$ where $Z$ is a countable infinite family of separated "holes". It's impossible to use a finite number of singletons to cover a infinite number of separated holes. $\Box$

What do you think about my proof? Who is right?

Answer 1

Concerning the current edition of your question, I can say that both propositions are right, by The Sierpiński Theorem. I recall that a continuum is a compact connected Hausdorff space and remark that the unit segment $[0,1]$ can contain at most a countable family of pairwise disjoint nondegenerate closed intervals.

(cited from “General Topology” by Ryszard Engelking. )

Answer 2

Your friend is right....(1): If it were possible it would need a countably infinite family...(2):Proof by contradiction: Suppose $I(n)=[A(n),,B(n)] $ with $A(n) < B(n)$ for $n \in N$ with $ \cup \{ I(n) \} _{n \in N} = [0,1]$ and $I(m) \cap I(n) = \phi$ for $m \not =n$. We may assume $A(0)=0$ and $B(1)=1$.For brevity,say that $I(j)$ is between $I(i)$ and $I(k)$ to mean that $$ \max I(i) < \min I(j) < \max I(j) < \min I(k)$$ or that $$\max I(k) < \min I(j) < \max I(j) < \max I(i).$$Now define $F(0)=0, F(1)=1$ and for $n \in N$ let $F(n+2)$ be the least $k > F(n+1)$ such that $I(k)$ is between $I(F(n)$ and $I(F(n+1)$. We have $$B(0)=B(F(0)) < B(F(2)) < B(F(4)) < ... < A(F(5)) < A(F(3)) < A(F(1))=A(1)$$ and F is strictly increasing... Now let $x$ be the lub of $ \{ B(F(2n)) \} _{n \in N}$. Then $x \in I(j)$ for some $j$, and $j$ not equal to any $F(n)$. So $F(n+1) < j < F(n+2)$ for some $n > 0$.(Recall $F(0)=0$ and $F(1)=1$.) But then $I(j)$ is between $I(F(n-1))$ and $I(F(n))$ with $F(N+1)) < j < F(n+2)$, contradicting the definition of $F(n+1)$. QED. Also observe that if we consider only countable families we may allow some $I(n)$ to be degenerate, with an obvious modification to my definition of "between".