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From Are the Sierpiński cardinal ˊn and its measure modification ˊm equal...?, I seem to have rediscovered a result from Sierpinski:

Theorem (Sierpiński, 1921). For any countable partition of the unit interval $[0,1]$ into closed subsets exactly one set of the partition is non-empty.

(1) Does the proof below work?

(2) It appears from How strong is Sierpiński theorem about continua? and Partitioning [0,1] into pairwise disjoint nondegenerate closed intervals that we can replace $[0, 1]$ with any continuum (nonempty compact connected Hausdorff space). Is there a complete proof of this on the web? The first answer to the second question above has most of a proof, but missing a lemma. (There's more discussion in Show [0,1]≠⋃∞𝑛=1𝐼𝑛....)

[EDIT: for general continua, in addition to the answer below, there's one in an answer to Is [0,1] a countable disjoint union of closed sets?. For the $[0, 1]$ result, there's also some nice discussion in Why are the integers with the cofinite topology not path-connected?]

Proof. Suppose not: then take just the nonempty closed sets, and the partition has size at least two. If the partition is finite, it disconnects $I = [0, 1]$. So the partition is countably infinite, say $(A_i)_0^\infty$. We'll get a contradiction by constructing a point of $I$ which is not in $\bigcup A_i$, by building a decreasing sequence of open intervals which stay away from the $A_i$.

For any $j$, take $B_j$ = $\bigcup_{i = 0}^j A_j$. $B_j$ is a closed proper subset of $I$.

Get $j_0$ with $0, 1 \in B_{j_0}$, and get $x_0 \in I - B_{j_0}$. Since $B_{j_0}$ is closed, there's an open interval $(t, u)$ disjoint from $B_{j_0}$ with $x_0 \in (t, u)$. Take $p_0$ to be the $\inf$ of all such $t$, and $q_0$ to be the $\sup$ of all such $u$: then $(p_0, q_0)$ is still disjoint from $B_{j_0}$. Note that $p_0 \in B_{j_0}$: otherwise $p_0$ has a neighborhood disjoint from $B_{j_0}$ and $p_0$ would not be minimal. Similarly, $q_0 \in B_{j_0}$.

Take the smallest index $j'_0$ such that $A_{j'_0}$ meets $(p_0, q_0)$ (note that $j'_0 > j_0$). I claim there is some $u \in (p_0, q_0)$ with $(p_0, u)$ disjoint from $A_{j'_0}$: otherwise there's a sequence $u_k \to p_0$ with all $u_k$ in $A_{j'_0}$, which is impossible since $A_{j'_0}$ is closed and does not contain $p_0$. Take $q_1$ to be the $\sup$ of all such $u$: then $(p_0, q_1)$ is disjoint from $A_{j'_0}$ and $q_1 \in A_{j'_0}$ as above. In fact $(p_0, q_1)$ is disjoint from $B_{j'_0}$, because $j'_0$ is minimal. Similarly, on the other side, take the smallest $j_1$ such that $A_{j_1}$ meets $(p_0, q_1)$, and take $p_1$ to be the $\inf$ of all $t \in (p_0, q_1)$ such that $(t, q_1)$ is disjoint from $A_{j_1}$. Now $j_1 > j_0$, $p_0 < p_1 < q_1 < q_0$, $(p_1, q_1)$ is disjoint from $B_{j_1}$, and $p_1, q_1 \in B_{j_1}$.

Repeat this to get $j'_1$ and $q_2$, and $j_2$ and $p_2$, with $p_1 < p_2 < q_2 < q_1$, $(p_2, q_2)$ disjoint from $B_{j_2}$, and $j_2 > j_1$. Continue inductively to get $j_i$ and $(p_i, q_i)$ for all $i$. The sequence $(j_i)$ approaches infinity, so $\bigcup B_{j_i} = I$. $(p_i)$ is increasing and $(q_i)$ is decreasing. Taking $p = \sup {p_i}$ and $q = \inf {q_i}$, we have $p \le q$, when means $p$ is in all the $(p_i, q_i)$. But then $p$ is in none of the $B_{j_i}$, which is impossible. Q.E.D.

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    $\begingroup$ You can find a proof for arbitrary continua as theorem 5.16 of Nadler's continuum theory book $\endgroup$ Commented Sep 27, 2023 at 10:32

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[edit: original post has now been modified accordingly, so this first section no longer applies]

Your proof needs to be modified slightly. In the first step you rely on the fact that $p_0 \in B_{j_0}$ in order to guarantee there is a suitable $q_1$ between $p_0$ and $q_0$, and similarly for $q_0$. You need this to be true for $p_i$ or $q_i$ in general each time you iterate through the process.

The easiest way to ensure this is so is to simply always choose the largest possible $q_i$, and likewise always choose the smallest $p_i$ (this is fine to do, as from your construction it is still clear that $p_i$ will be strictly increasing and $q_i$ will be strictly decreasing). Other than that your proof looks fine to me.


As for the general theorem for continua, I don't know of an online source unfortunately, though I do I see Nadler's book has been mentioned in the comments.

However, I believe I can infer what the missing lemma is in the proof in Engelking you linked, so here that is:


Lemma 1. Let $X$ be a continuum and let $A\subset X$ be a proper closed subset. Then every component of $A$ intersects $\partial A$.

To prove this, suppose otherwise and let $C\subseteq A^o$ be a component. Then for each $x\in \partial A$ we may choose open sets $U_x,V_x\subseteq X$ with $U_x\cup V_x\supseteq A$, $U_x\cap V_x\cap A=\emptyset$, $U_x\supseteq C$, $V_x\ni x$. By compactness there's a finite set $S\subseteq \partial A$ with $$V:=\bigcup_{x\in S} V_x\supseteq \partial A\text{,}$$ and $$U:=\bigcap_{x\in S} U_x \supseteq C\text{,}$$ still open, and satisfying $U\cup V\supseteq A$ and $U\cap V\cap A=\emptyset$.

Since $V\supseteq \partial A$ and $U\supseteq C$, we see $U\cap A^o$ and $V\cup X\backslash A$ are nonempty open sets, with $(U\cap A^o)\cup (V\cup X\backslash A) = X$, and $(U\cap A^o)\cap (V\cup X\backslash A) =\emptyset$, contradicting connectedness of $X$.


From the above lemma you can patch the proof in Engelking linked above.

It might be nice to have a self-contained proof online in one place, and it seems your proof idea has a slightly more intuitive flavor than the one in Engelking (though ultimately the same principles are at work).

I'd therefore like to show how to generalize the main idea from your own method to prove the theorem. The key ideas are the same as the proof in Engelking, but presented in a way more in analogy with your proof above.

I don't know if this exact presentation is done elsewhere, though it wouldn't surprise me at all if it was.


Preliminaries

We will use a corollary to the preceding lemma (which captures the same idea as Lemma 6.1.26 preceding the linked Engelking proof.)

Lemma 2. Let $Z$ be a continuum and let $A,B\subset Z$ be disjoint proper closed subsets, with $A\neq\emptyset$. Then there is a continuum $C\subseteq Z$ intersecting $A$ and $Z\backslash A$ and disjoint from $B$.

To prove this, by normality let $U\supseteq A$ be open with $\bar{U}\cap B=\emptyset$. Let $x\in A$, and let $C$ be the component of $x$ in $\bar{U}$. By Lemma 1, $$C\cap Z\backslash A\supseteq C\cap \partial \bar{U}\neq \emptyset\text{,}$$ and of course clearly $C$ intersects $A$ and not $B$.


Proof of Sierpiński's theorem for continua.

Let $X$ be a continuum, and define the partition $A_i$ of $X$ and the increasing union $B_j=\bigcup_{i=i}^j A_i$ as in your original proof.

Let $X_1=X$, $j_1=1$.

Observe that for $i=1$, we have

  1. $X_i$ is a continuum, with $X_k\supseteq X_i$ for each $k<i$ (vacuously so far).
  2. $X_i\cap B_{j_k}=\emptyset$ for every $k<i$ (vacuously so far).
  3. $X_i\cap B_{j_i}$ is a proper nonempty subset of $X_i$.

Having defined $X_i$ and $j_i$ to satisfy the preceding, we let $j_{i+1}$ be the first index for which $X_i\cap B_{j_{i+1}}\backslash B_{j_i}\neq \emptyset$.

Now let $A=X_i\cap B_{j_{i+1}}\backslash B_{j_i}$, $B=X_i\cap B_{j_i}$, and $Z=X_i$, and apply Lemma 2 to obtain some continuum $X_{i+1}\subseteq X_i$ intersecting $A$ and $X_i\backslash A$ and disjoint from $B$. In particular, $X_{i+1}\cap B_{j_{i+1}}$ is a proper nonempty subset of $X_{i+1}$, and $X_{i+1}\cap B_{j_i}=\emptyset$.

Hence $X_{i+1}$, and $j_{i+1}$ continue to satisfy the three conditions.

By induction the sequence is well-defined. But we have a contradiction, since by compactness, the intersection $\bigcap_i X_i$ is nonempty, but by the second condition, $\bigcap_i X_i$ must be empty.

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  • $\begingroup$ Thank you, that all looks great but will take me some time to absorb. $\endgroup$
    – Hew Wolff
    Commented Sep 30, 2023 at 13:28
  • $\begingroup$ OK, I think I fixed up my proof. $\endgroup$
    – Hew Wolff
    Commented Oct 14, 2023 at 15:31
  • $\begingroup$ @HewWolff just looked over it, it looks good, I believe everything is right now. $\endgroup$
    – M W
    Commented Oct 15, 2023 at 4:15
  • $\begingroup$ In your Lemma 2, I think technically you should also mention the case that $B$ is empty, because you need $B$ to be nonempty to guarantee that $\bar{U}$ is proper so you can use Lemma 1. This case is trivial because we just choose $C = Z$. $\endgroup$
    – Hew Wolff
    Commented Dec 7, 2023 at 17:31

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