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Let $X$ be a continuum $=$ a connected compact metric space.

Define $x\sim y$ if $x$ and $y$ are contained in a nowhere dense subcontinuum of $X$. It is easy to see that $\sim$ is an equivalence relation.

Examples:

$|[0,1]/\sim|=|\mathbb R|$

$|[0,1]^2/\sim|=1$

Question: Is there a continuum $X$ with $1<|X/\sim|<|\mathbb R|$?

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  • $\begingroup$ Being pedantic: $X \ne \{*\}$ $\endgroup$ – user60589 Nov 21 '16 at 11:50
  • $\begingroup$ Although I don't have an idea how to answer your question, it seems that you should reformulate your question: What is the cardinality of $X/\sim$? If it is infinite then it may be countable or uncountable. At the moment I only see uncountable examples (leaving aside the continuum hypothesis). $\endgroup$ – Paul Frost Jun 1 '18 at 23:02
  • $\begingroup$ @PaulFrost Done. $\omega$-many classes would be interesting as well. $\endgroup$ – Forever Mozart Jun 1 '18 at 23:12
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This is an incomplete answer (this question is non-trivial, it is actually quite difficult IMO) but I thought I'd put it here with the hope that someone could complete the proof (I've spent a good amount of time on it already and unfortunately have to move on).

$\textbf{Proposition:}$ Suppose $X$ is a path-connected continuum. Then under the equivalence relation previously described, there are either infinitely many equivalence classes or there is only one.

$\textbf{Proof:}$ There are two cases here:

1.) If for each $x \neq y \in X$, there is a path $f: [0,1] \rightarrow X$ between $x$ and $y$ that is nowhere dense in $X$, then the closure $\overline{f([0,1])} \subseteq X$ is a nowhere dense, connected (closure of continuous image of connected set), compact (closed subset of compact space) set and hence a nowhere dense subcontinuum. Thus, in this case, we have $x \sim y$ for each pair $x \neq y$ and as a result only one equivalence class.

2.) Now suppose we have $x \neq y \in X$ such that every path $f:[0,1] \rightarrow X$ between them is dense in some nonempty open set. As a metric space, $X$ is Hausdorff and so path connectedness implies arc-connectedness, so there is a path between $x$ and $y$, $f:[0,1] \rightarrow f([0,1]) \subseteq X$, that is also a homeomorphism. By hypothesis, there is a non-empty open ball $U \subseteq X$ such that $f([0,1]) \cap U =V$ is dense in $U$. We now argue that $U \cong (a,b) \subseteq \mathbb{R}$ for some $a<b$.

Consider $U$ as a subspace of $X$. Note that $f([0,1]) \subseteq X$ is closed and thus by definition of the subspace topology, $V$ is closed. We then define the map $F: U \rightarrow \mathbb{R}$, $F: x \mapsto d(x, V)$, where $d$ is the metric defined on $X$. Since $V$ is closed, $F$ is continuous. But note that for each $x \in V$, $F(x)=0$, so $F$ agrees with the zero function on a dense subset of a Hausdorff space and so must be $0$ everywhere. Then $F(U)= \left\{0 \right\}$, i.e. for every $x \in U$, $d(x, V)=0$. But then $U=V$. Then since $U$ is an open ball, it is connected and its image under the homeomorphism $f^{-1}: f([0,1]) \rightarrow [0,1]$ must be connected and open, and so is an interval of the form $(a,b)$. Then $U \cong (a,b)$ as claimed. (Obviously, $U \neq \emptyset$ implies $a<b$)

$\textit{Lemma:}$ Suppose $S \subseteq X$ is a connected set such that there are two distinct points $c,d \in S \cap U$. Then $S \cap U$ is connected.

(presented without proof, I haven't been able to prove this lemma).

If we suppose we have $c,d \in U$ such that $c\sim d$, then there is a connected, nowhere dense set $S$ with $c,d \in S$. Then by the lemma, $S \cap U$ is connected and since it contains at least two points, its image under the homeomorphism $f^{-1}$ is an interval. But then $S$ is dense in some open set in $U$, contradicting the "nowhere dense" hypothesis. We then conclude that given two points $c,d \in U$, they cannot be elements of the same equivalence class. It then follows that there are infinitely many classes, since $U$ has infinitely many elements. This completes the proof. $\Box$

So it comes down to that lemma (assuming everything else is OK). It seems intuitively true but I haven't actually been able to prove it. Also, I can't see any immediate way to extend this to the case where the space is just connected.

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  • $\begingroup$ The arc attached to the square actually illustrates the difficulty with proving that lemma quite nicely. In this case, we have an open ball homeomorphic to a line segment, which takes the role of $U$ in the proof above. How do we prove that any connected subset containing two elements of the arc must have connected intersection with the arc? $\endgroup$ – M10687 Nov 26 '16 at 20:17
  • $\begingroup$ To prove lemma, it suffices to show that $S\subseteq U$. Well, let $s\in S\cap U$ and suppose $S\setminus U\neq\varnothing$. Since $S$ is connected we must have $a\in S$ or $b\in S$. Assuming $a\in S$, then we have $[a,s]\subseteq S$, whence $S$ has nonempty interior. $\endgroup$ – Forever Mozart Nov 26 '16 at 21:57
  • $\begingroup$ Actually I think the Lemma as stated is false because $S$ could go into either side of $[a,b]$ but not all the way through. But my last comment is sufficient for your argument. Oh, and also, the closure of $(a,b)$ might not be an arc. However, remainder of the closure of $(a,b)$ has at most 2 components, and so my argument should still work if you replace $a$ with the limiting continuum at that end... $\endgroup$ – Forever Mozart Nov 27 '16 at 1:45
  • $\begingroup$ I fail to see how that would imply the Lemma is false. However, seeing that the previous comment suffices to complete the proof for path-connected spaces, we don't really need that lemma anymore. I'm going to edit the answer to reflect your comment. $\endgroup$ – M10687 Nov 27 '16 at 1:55
  • $\begingroup$ Yeah the important thing is that if you go from a limiting end of $(a,b)$ into $(a,b)$ with a connected set, then you must include an interval of $(a,b)$. $\endgroup$ – Forever Mozart Nov 27 '16 at 1:59
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This is not an answer, but perhaps helps to understand the nature of possible examples.

Let us extend $\sim$ to arbitrary compact metric spaces $X$ by defining

$x \sim y$ if $x = y$ or $x$ and $y$ are contained in a nwd (= nowhere dense) subcontinuum of $X$.

If $X$ does not have isolated points (which is the case if $X$ is a continuum with more than one point), then the part $x = y$ can be omitted.

Let us denote the equivalence classes rel. $\sim$ as "nwd-components''. We define

$$\nu(X) = \lvert X/\sim \rvert .$$

We shall prove

Theorem. The following are equivalent:

(1) There exists a continuum $X$ such that $\nu(X) = 2$.

(2) For each $n \ge 2$ there exists a continuum $X$ such that $\nu(X) = n$.

(3) There exist $n \ge 2$ and a continuum $X$ such that $\nu(X) = n$.

Moreover, if one of these is satisfied the following holds:

(4) There exists a continuum $X$ such that $\nu(X) = \aleph_0$.

We begin with

Lemma. Let $X$ be a compactum, $x_1,x_2 \in X$ be two distinct points and $p : X \to X' = X/\lbrace x_1, x_2 \rbrace$ be the quotient map (note that $X'$ is also a compactum and $p$ a closed map). Let $A_i$ denote the nwd-components of $x_i$ in $X$ and $A'$ the nwd-component of $\ast$ in $X'$, where $\ast = p(x_1) = p(x_2)$. If $A_1 = A_2$, then $p$ establishes a bijection between the nwd-components of $X$ and those of $X'$; otherwise $p(A_1) \cup p(A_2) = A'$ and $p$ establishes a bijection between the nwd-components of $X$ different from $A_1, A_2$ and the nwd-components of $X'$ different from $A'$ (that is, two nwd-components of $X$ are merged to one nwd-component of $X'$).

Proof. If $x_1$ or $x_2$ is an isolated point, then this trivial. So let us assume that $x_1,x_2$ are non-isolated points.

a) Observe that if $U \subset X$ is open, then $p(U) \subset X'$ is open iff either $x_1,x_2 \in U$ or $x_1,x_2 \notin U$.

b) Let $N' \subset X'$ be closed and $N = p^{-1}(N')$. Then $N' \subset X'$ is nwd iff $N \subset X$ is nwd.

Assume $U = int(N) \ne \emptyset$. We have $V = U \backslash \lbrace x_1, x_2 \rbrace \ne \emptyset$, because otherwise $U \subset \lbrace x_1, x_2 \rbrace$ which would imply that at least one of $x_1,x_2$ is an isolated point. Therefore $p(V)$ is open in $X'$ and contained in $N' = p(N)$.

Assume $U' = int(N') \ne \emptyset$. Then $p^{-1}(U')$ is open in $X$ and contained in $p^{-1}(N') = N$.

c) Let $C' \subset X'$ be closed and $C = p^{-1}(C')$. Then $C'$ is connected iff either $C$ is connected or $C = C_1 \cup C_2$ with two disjoint closed connected $C_i$ such that $x_i \in C_i$.

One half is trivial since $p(C) = C'$. Assume $C'$ is connected. If $\ast \notin C'$, then $C$ and $C'$ are homeomorphic and $C$ is connected. So let $\ast \in C'$. We only consider the non-trivial case that $C$ is not connected. Then there are two disjoint nonempty closed $C_i$ such that $C = C_1 \cup C_2$ and $x_1 \in C_1$. We must have $x_2 \in C_2$, otherwise $p(C_1), p(C_2)$ would be disjoint nonempty closed subsets of $C'$ whose union is $C'$. Assume that $C_1$ is not connected. Then there are two disjoint nonempty closed $D_j$ such that $C_1 = D_1 \cup D_2$ and $x_1 \in D_1$. Then $E_1 = D_1 \cup C_2$ and $D_2$ are disjoint nonempty closed subsets of $C$ and $p(E_1), p(D_2)$ are disjoint nonempty closed subsets of $C'$ whose union is $C'$, a contradiction. Hence $C_1$ must be connected, and similarly also $C_2$.

We now completely understand the nwd subcontinua $C'$ of $X'$: Any such $C'$ is either the image $p(C)$ of a nwd subcontinuum $C$ of $X$ or the union of two images $p(C_i)$ with disjoint nwd subcontinua $C_i \ni x_i$. The nwd-part ist true because subspaces of nwd subspaces are again nwd and the union of two closed nwd subspaces is nwd.

This yields the lemma.

Proof of Theorem.

$(1) \Rightarrow (2)$: Let $X$ be a continuum with $\nu(X) = 2$. Pick any $x \in X$, take the disjoint union $X + X$ and identify the two copies of $x$. This yields $X'$ with $\nu(X') = 3$. Proceed inductively to reach an arbitrary $n$.

$(2) \Rightarrow (3)$: Trivial.

$(3) \Rightarrow (1)$: Use the Lemma to reduce the number of nwd components by identifying points until you end with two nwd components.

$(1) \Rightarrow (4)$: Given a sequence of pointed spaces $(X_j,x_j)$, let us define the cluster $$X_\infty = cluster_{j=1}^\infty (X_j,x_j) = \lbrace (z_j) \in \Pi_{j=1}^\infty X_j \mid z_j \ne x_j \text{ for at most one } j \rbrace$$ with the subspace topology inherited from the product. The point $\ast = (x_j)$ is the canonical basepoint of the cluster. There are canonical embeddings $X_k \to X_\infty$ and by an abuse of notation we write $X_k \subset X_\infty$. Doing so we have $x_k = \ast$. $X_\infty$ is the disjoint union of $\ast$ and the sets $X_k \backslash \ast$. Any $Z \subset X_\infty$ can be written as $Z = \bigcup_{j=1}^\infty Z_j$ with $Z_j = Z \cap X_j$ ("canonical decomposition''). If $\ast \in Z$, then each $Z_j$ is a retract of $Z$ (all $z \notin Z_j$ can be mapped to $\ast \in Z_j$). A basis of open neighborhoods of $\ast$ in $X_\infty$ is given by the sets $\bigcup_{j=1}^\infty U_j$ with open neighborhoods $U_j$ of $x_j$ in $X_j$ such that $U_j = X_j$ for $j \ge m$. If $z \in X_j \backslash \ast$ has an open neighborhood $U$ not containing $\ast$, then a basis of open neighborhoods of $z$ in $X_\infty$ is given by the open subsets of $U$ in $X_j$ which contain $z$. If $x_k$ is closed in $X_k$, then $X_k \backslash \ast$ is open in the cluster.

Let $X$ be a compactum with $\nu(X) =2$, i.e. having two nwd-components $A, B$. Pick $x \in A$. Let $X_\infty = cluster_{j=1}^\infty (X_j,x_j)$ where the $(X_j,x_j)$ are copies of $(X,x)$. Their two nwd-components are $A_j$ and $B_j$. $X_\infty$ is compact because every open neighborhood of $\ast$ contains all but finitely many $X_j$. Let $C$ be a subcontinuum of $X_\infty$.

Case 1: $\ast \notin C$. Then $C$ cannot intersect more than one $X_k\backslash \ast$ since these sets are open in $X$; hence $C \subset X_k \backslash \ast$ for a unique $k$. Since $X_k \backslash \ast$ is open in $X_\infty$, we see that $C$ is nwd in $X_\infty$ iff it is nwd in $X_k$.

Case 2: $\ast \in C$. Let $C = \bigcup_{j=1}^\infty C_j$ be the canonical decomposition. The $C_j$ are subcontinua of $X_j$ because they are retracts of $C$. We have $\ast \in C_j$. We show that $C$ is nwd in $X_\infty$ iff all $C_j$ are nwd in $X_j$.

a) If all $C_j$ are nwd in $X_j$, then $C$ is nwd in $X_\infty$: Assume that $z$ is an interior point of $C$. If $z \ne \ast$, then $z \in X_k \backslash \ast$ for some $k$ so that it would be an interior point of $C_k$. If $z = \ast$, there are open neighborhoods $U_j \subset X_j$ of $x_j$, $U_j = X_j$ for $j \ge m$, such that $\bigcup_{j=1}^\infty U_j \subset C$. Hence $U_j \subset C_j$ so that the $\ast$ would be an interior point of each $C_j$.

b) If some $C_k$ has an interior point $z_k$ in $X_k$, then $C$ has an interior point in $X_\infty$: (i) If $z_k \ne \ast$, then $z_k$ is an interior point of $C$. (ii) If $z_k = \ast$, choose an open $U_k \subset X_k$ with $\ast \in U_k \subset C_k$. $U_k$ must contain a point $z_k \ne \ast$, otherwise $\ast$ would be open in $X_k$ so that $X_k$ cannot be connected (note that (1) implies that $X$ has more than one point). This reduces (ii) to (i).

This immediately implies that the nwd-components of $X_\infty$ are $\bigcup_{j=1}^\infty A_j$ and the $B_j$ which are countably many.

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  • $\begingroup$ It is fairly obvious to me that $(3)\Rightarrow(1)$, but I like the construction for $(1)\Rightarrow(4)$. I am trying to prove $(4)\Rightarrow(1)$ at the moment. $\endgroup$ – Forever Mozart Jun 11 '18 at 18:08

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