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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.

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  • $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Apr 12 at 19:53
  • $\begingroup$ Should it be obvious that if $X$ is Suslinean then so is $X / \sim$? $\endgroup$ – Will Brian Apr 12 at 20:01
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    $\begingroup$ @WillBrian Maybe not in general, but the quotient map here is monotone so that should do it. $\endgroup$ – D.S. Lipham Apr 12 at 20:03
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    $\begingroup$ Why is there a vote to close? $\endgroup$ – D.S. Lipham Apr 13 at 22:50
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    $\begingroup$ At this point, you might want to look at mathoverflow.net/questions/31337/…. $\endgroup$ – Will Brian Apr 15 at 13:18
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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.

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