# A continuum which is both Suslinean and non-Suslinean?

Continuum means compact connected metrizable with more than one point.

A continuum is Suslinean if every collection of pairwise disjoint subcontinua is countable.

There is an apparent contradiction in the literature I would like to resolve...

In Example 3 at the end of Paper A, there is constructed a continuum $$Y:=X/\sim$$ which is the quotient of another continuum $$X$$. It is clear from the construction that $$X$$ is Suslinean, and since the quotient map is monotone $$Y$$ is also Suslinean. Actually, the author claims $$X$$ and $$Y$$ are hereditarily locally connected, and such continua are automatically known to be Suslinean.

On the other hand, $$Y$$ is the closure of a ray (a one-to-one continuous image of $$[0,1)$$) such that the ray is first category in $$Y$$. In other words, the ray and its complement are each dense in $$Y$$. It follows that there is a sequence of pairwise disjoint arcs in $$Y$$ which converges to $$Y$$ in the Hausdorff distance. By Theorem 30 in Paper B, $$Y$$ is non-Suslinean.

Question 1: Am I correct in finding a contradiction?

Question 2: Is Example 3 correct? Specifically, why is it Hausdorff (and therefore metrizable)? I found a couple of typos, e.g. $$A_n$$ should be $$C_n\cup \bigcup ...$$ and $$\overline{z_1 z_1}$$ should be $$\overline{z_1 z_2}$$, but otherwise it seems okay.

If there is a problem with Theorem 30, then I think the error must be in the proof of Lemma 29. Specifically there is a claim that $$H_W$$ is non-degenerate since it is an inverse limit of non-degenerate continua. This is a false statement in general because we can take the continua $$[0,1/n]$$ with bonding maps the inclusions $$[0,1/(n+1)] \hookrightarrow[0,1/n]$$, and the inverse limit is just the single point $$\langle 0,0,..\rangle$$. However, since each factor of the inverse limit for $$H_W$$ projects into $$\partial U$$ and $$\partial V$$ in the first factor, the inverse limits on preimages of these boundary sets should be non-empty and disjoint subsets of $$H_W$$...

Upon closer inspection, the proof of Lemma 29 (Paper B) is flawed in more ways than one. I'm not sure it can be saved.

At this point I would lean toward the example being correct. I really just need to see that $$Y=X/\sim$$ is Hausdorff.

Paper A:

Tymchatyn, E. D., Some rational continua, Rocky Mt. J. Math. 13, 309-319 (1983). ZBL0514.54022.

Paper B:

Mouron, Christopher, The topology of continua that are approximated by disjoint subcontinua, Topology Appl. 156, No. 3, 558-576 (2009). ZBL1165.54012.

• Good question, bad title. I suggest something with Suslinean in the title, e.g. "To Be Suslinean Or Not; That Is The Question". Gerhard "Brush Up On Your Shakespeare" Paseman, 2019.04.12. – Gerhard Paseman Apr 12 at 19:53
• Should it be obvious that if $X$ is Suslinean then so is $X / \sim$? – Will Brian Apr 12 at 20:01
• @WillBrian Maybe not in general, but the quotient map here is monotone so that should do it. – D.S. Lipham Apr 12 at 20:03
• Why is there a vote to close? – D.S. Lipham Apr 13 at 22:50
• At this point, you might want to look at mathoverflow.net/questions/31337/…. – Will Brian Apr 15 at 13:18

Example 3 in Paper A is indeed a counterexample to Theorem 30 in Paper B.

Lemma 29 must also be false because it implies Theorem 30.

The only thing left to verify is that $$Y=X/\sim$$ (from Example 3) is Hausdorff, so that $$Y$$ is actually a metrizable continuum. This can be proved in a few steps:

1. For every compact $$K\subseteq [0,1]$$ the set $$\widehat{K}:=K\cup \bigcup \big\{\overline{z_1 z_2}:z_1\text{ and }z_2\text{ are consecutive endpoints of some }C_n \text{ and }(z_1\in K\text{ or }z_2\in K)\big\}$$ is a closed (compact) subset of $$X$$.
2. If $$U$$ is any open subset of $$X$$ containing an element $$y\in Y$$, then $$V=X\setminus\widehat{[0,1]\setminus U}$$ is an open subset of $$X$$ which is a union of members of $$Y$$, and $$y\subseteq V\subseteq U$$.
3. $$Y$$ is Hausdorff by 2 and normality of $$X$$.

We can argue 1 as follows. Let $$x$$ be any point in the closure of $$\hat K$$. Suppose for a contradiction that $$x\notin \hat K$$. Then $$x\in [0,1]\setminus K$$ and there is a sequence of semi-circular arcs $$\overline{z^n_1 z^n_2}\in Y$$ with $$z^n_1\in K$$ and $$z^n_2\to x$$. (I assume without loss of generality that the subscripts on the $$z$$'s are arranged in this way).

Since $$d(x,K)>0$$ this means infinitely many semicircles have radius greater than some positive constant. But this is impossible by the construction of $$X$$. Therefore $$x\in \hat K$$ and $$\hat K$$ is compact.

Now 1 $$\Rightarrow$$ 2 $$\Rightarrow$$ 3 and the proof is complete.