Continuously selecting elements from unordered pairs

Question

The symmetric square of a topological space $X$ is obtained from the usual square $X^2$ by identifying pairs of symmetric points $(x_1,x_2)$ and $(x_2,x_1)$. Thus, elements of the symmetric square can be identified with unordered pairs $\{x_1,x_2\}$ from $X$ (including the degenerate case $x_1 = x_2$). With this identification, the following question is very natural:

When is there a continuous selector from the symmetric square to the original space?

This is always possible if $X$ is a generalized ordered space (a subspace of a linear order with the order topology) since min and max are both continuous selectors. I don't think this is a necessary condition since I can construct continuous selectors (on Baire space, for example) that aren't equal to min or max for any ordering of the underlying space.

A necessary condition is that removing the diagonal disconnects $X^2$. More precisely, as Adam Bjorndahl pointed out in the comments, the complement of the diagonal should be the union of two disjoint open sets such that $(x_1,x_2)$ is in one iff $(x_2,x_1)$ is in the other. So, for example, there are no continuous selectors for the symmetric square of the plane nor the circle.

I'm mostly interested in the case when $X$ is Polish, so it's fine to assume that $X$ is very nice: Hausdorff, normal, perfect, etc.

Answer 1

In this paper Van Mill and Wattel proved that the existence of a continuous selection characterizes orderability in the class of compact Hausdorff spaces.

Answer 2

This should have been a comment, but it got a bit too long.

A possibly useful necessary condition, in regular spaces, for the existence of a continuous selector is that there should not exist three points $a,b,c\in X$ such that each two lie in a closed connected set that misses the third point. Proof: If we had three such points and closed connected sets, then we could argue as follows for any alleged selector $S$. Suppose, without loss of generality, that $S\{a,b\}=a$. Let $M$ be a closed connected set containing $b$ and $c$ but not $a$. By regularity of $X$, $a$ and $M$ have disjoint open neighborhoods. So if we let $m$ vary over $M$, the values of $S\{a,m\}$ form a connected subset of $M\cup\{a\}$ that contains $a$, which means none of these values can be in $M$ because of the disjoint neighborhoods. In particular, $S\{a,c\}=a$. Now hold $c$ fixed and consider $S\{n,c\}$, where $n$ varies over a closed connected set $N$ that contains $a$ and $b$ but not $c$. An argument like the preceding one shows that $S\{n,c\}$ must stay in $N$, so in particular $S\{b,c\}=b$. A third use of the same argument, holding $b$ fixed and moving $c$ to $a$ in a closed connected set that misses $b$ shows that $S\{b,a\}=b$, contrary to our initial assumption.

This implies in particular that, if $X$ is to have a continuous selector from pairs, then $X$ cannot contain a triod or a circle (homeomorphic copies of the letters F and D). If we were dealing with CW complexes, we could deduce that the dimension is at most 1 (so we have a graph), that the graph must be acyclic (so we have a forest), and that there must be no branching (so we have a disjoint union of real intervals). For not-so-nice spaces, the situation is unfortunately not so clear.

Answer 3

Like the first answer, this is really a comment that has outgrown the comment box. No new results here, but a different way of looking at continuous selectors that may be helpful (somewhat analogous to the relationship between choice functions and preference relations in decision theory).

Call a relation $T$ on a topological space $X$ an open tournament on $X$ (thanks for the terminology!) if the following conditions hold:

(a) $T$ is total: for all $x,y \in X$, either $xTy$ or $yTx$;

(b) $T$ is antisymmetric: for all $x,y \in X$, if $xTy$ and $yTx$ then $x = y$;

(c) $T$ respects the topology on $X$: for all distinct $x,y \in X$ with $xTy$, there exist disjoint open neighbourhoods $U$ and $V$ about $x$ and $y$, respectively, such that for all $x' \in U$ and all $y' \in V$, $x'Ty'$.

Let's read $xTy$ and "$x$ trounces $y$" (yes, I am having some fun with this).

Lemma: The existence of a continuous selector for a Hausdorff space $X$ is equivalent to the existence of an open tournament on $X$.

Proof. Given an open tournament $T$ on $X$, define $s: [X]^{2} \to X$ by $$s(\{x,y\}) = x \iff xTy.$$ Conditions (a) and (b) ensure that $s$ is a well-defined selector function. Moreover, if $s(\{x,y\}) = x$ and $W$ is an open neighbourhood about $x$, then since $xTy$ we can find $U$ and $V$ as in condition (c), in which case the open sets $U \cap W$ and $V$ yield an open neighbourhood about $\{x,y\}$ contained in $s^{-1}(W)$, so $s$ is continuous. Note also that conditions (a) and (c) imply that $X$ is Hausdorff.

Conversely, suppose that $X$ is a Hausdorff space and $s$ is a continuous selector function. Define $T \subset X \times X$ by: $$xTy \iff s(\{x,y\}) = x.$$ It is clear that $T$ is total and antisymmetric. Suppose that $x \neq y$ and $xTy$, so we have $s(\{x,y\}) = x$. Let $W$ and $W'$ be disjoint neighbourhoods of $x$ and $y$, respectively, and let $U$ and $V$ be the open neighbourhoods of $x$ and $y$ corresponding to $s^{-1}(W)$. Then $W \cap U$ and $W' \cap V$ are as required in condition (c). $\blacksquare$

Given an open tournament $T$ on $X$, for each $x \in X$ define $$T_{x} := \{y \in X : x \neq y \textrm{ and } xTy\}$$ and $$T^{x} := \{y \in X : x \neq y \textrm{ and } yTx\}.$$ Then it is easy to see that for all $x \in X$, $T_{x}$ and $T^{x}$ are open, and moreover $$T_{x} \sqcup T^{x} = X - \{x\}.$$ This shows that a necessary condition for the existence of an open tournament on $X$ is that $X$ is disconnected whenever it is punctured. In particular, this provides another way of seeing that neither $\mathbb{R}^{2}$ nor $S^{1}$ admit continuous selectors. This condition is certainly not sufficient, however, since it is satisfied by the triod. In that case, something more subtle seems to be going on; loosely speaking, the problem seems to be that distinct punctures in a neighbourhood of the vertex create very different separations.

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Edit: Let me try to make this last point a bit more precise. First some intuition. If $A$ and $B$ constitute a separation of a space $Y$, then $\chi_{A}: Y \to \{0,1\}$, the characteristic function of $A$, is both surjective (since $A \neq \emptyset$ and $A \neq Y$), and also continuous. Conversely, every continuous surjective characteristic function $\chi$ on $Y$ yields the separation $\chi^{-1}(1) \sqcup \chi^{-1}(0)$ of $Y$.

The product topology on the set $2^{Y}$ tells us that $\chi$ and $\chi'$ are "$F$-close" if they agree on the finite set $F \subset Y$ (these are the basic opens). This in turn provides a notion of closeness on the set $$\mathcal{S}_{Y} := \{(A,B) : \textrm{$A$ and $B$ constitute a separation of $Y$}\}$$ of all (ordered) separations of $Y$: $(A,B)$ and $(A',B')$ are "$F$-close" if $A \cap F = A' \cap F$ (so also $B \cap F = B' \cap F$).

More generally, given a fixed Hausdorff space $X$, let $$\mathcal{D} = \mathcal{D}_{X} := \{(A,B) : \textrm{$A$ and $B$ are open and disjoint}\},$$ and topologize this set using the notion of "$F$-closeness" as above: given a finite set $F \subset A \cup B$, say that $(A,B)$ and $(A',B')$ are "$F$-close" if $A \cap F = A' \cap F$ and also $B \cap F = B' \cap F$.

To connect this reasoning to the question at hand, observe that an open tournament $T$ on $X$ yields a function $f: X \to \mathcal{D}$ by setting $f_{T}(x) = (T_{x}, T^{x})$.

Lemma: $f_{T}$ is continuous.

Proof. Consider the basic open neighbourhood of $f_{T}(x) = (T_{x}, T^{x})$ corresponding to the finite set $F \subset T_{x} \cup T^{x} = X - \{x\}$. If $z \in T_{x} \cap F$, then in particular $xTz$ so there are open sets $U_{z}$ and $V_{z}$ about $x$ and $z$ respectively as in condition (c). Similarly, if $z \in T^{x} \cap F$ then $zTx$ so there are open sets $U_{z}$ and $V_{z}$ about $z$ and $x$ respectively as in condition (c). Set $$W := \bigcap_{z \in T_{x} \cap F} U_{z} \cap \bigcap_{z \in T^{x} \cap F} V_{z}.$$ Then $W$ is an open neighbourhood of $x$ and for any $y \in W$ we have $$T_{y} \cap F = T_{x} \cap F \textrm{ and } T^{y} \cap F = T^{x} \cap F,$$ which shows that $f$ is continuous. $\blacksquare$

Now we can see (from another perspective) why the triod admits no open tournament $T$: if it did and $v$ is the vertex point of the triod, then either $T_{z}$ consists of two of the prongs and $T^{z}$ consists of the third, or vice versa. However, in either case, $f_{T}$ cannot be continuous, since by choosing a point $w$ very close to $v$ (but distinct from $v$), we can always ensure that any two prongs lie in the same connected component, so we can always "ruin" whatever arbitrary choice was made at the vertex by $T$.

Here is a quick attempt at a converse. Let $f: X \to \mathcal{D}$ satisfy:

• $f$ is continuous;
• for each $x \in X$, $f(x) \in \mathcal{S}_{X - \{x\}}$;
• for all $x,y \in X$, $y \in (\pi_{0} \circ f)(x)$ iff $x \notin (\pi_{0} \circ f)(y)$.

Call such an $f$ (provisionally) "nice". One can check that when $T$ is an open tournament, $f_{T}$ is "nice". Define a binary relation $T_{f}$ on $X$ by: $$xT_{f}y \iff y \in (\pi_{0} \circ f)(x).$$ Then I believe/hope that $T_{f}$ is an open tournament (though I still need to check this carefully).

A converse like this would be appealing because it provides a nice picture of when there exists a continuous selector. For example, $\mathbb{R}$ would have one due to the "nice" function $$f(x) = (\{y : y < x\}, \{y : y > x\}).$$ Perhaps of more interest, the space $\mathbb{Q}^{2}$ would admit a continuous selector based on the existence of the following "nice" function: for each $p \in \mathbb{Q}^{2}$, let $L_{p}$ denote the subset of $\mathbb{Q}^{2}$ to the left of the line through $p$ of slope $\pi$, let $R_{p}$ denote the subset of $\mathbb{Q}^{2}$ to the right of the line through $p$ of slope $\pi$, and set $f(p) = (L_{p}, R_{p})$.

Answer 4

This is an expansion on the comment I made to the effect that the symmetric square of $\mathbb{Q}^2$ has a continuous selector, which is surprising since the symmetric square of $\mathbb{R}^2$ does not have a continuous selector.

In fact, a much more general statement is true:

The symmetric square of a countable metric space always has a continuous selector.

Let $X$ be a countable metric space. I will show that $X^2$ minus the diagonal can be split into two open sets $U$ and $V$ such that $(x,y) \in U$ iff $(y,x) \in V$. This is sufficient since $T = X^2 - U$ is then an open tournament in the sense of Adam Bjorndahl.

Let $d$ be a metric on $X^2$ such that $(x,y) \mapsto (y,x)$ is an isometry (e.g., let $$d((x_1,y_1),(x_2,y_2)) = d_0(x_1,x_2) + d_0(y_1,y_2)$$ where $d_0$ is a metric on $X$). Since $X$ is countable, the set $D$ of all possible values of $d$ is also countable, which means that the set $E = (0,\infty) - D$ of non-values of $d$ contains arbitrarily small positive numbers.

Fix an enumeration $(x_0,y_0),(x_1,y_1),\ldots$  of $X^2$ minus the diagonal. We will define $U$ and $V$ by stages.

To start things off, let $\varepsilon_0 \in E$ be sufficiently small that the open ball $B_{\varepsilon_0}(x_0,y_0)$ does not intersect the symmetric ball $B_{\varepsilon_0}(y_0,x_0)$. (In particular, neither ball intersects the diagonal of $X^2$.) Put all points of $B_{\varepsilon_0}(x_0,y_0)$ in $U$ and all points of $B_{\varepsilon_0}(y_0,x_0)$ in $V$.

Next, we consider the point $(x_1,y_1)$. If $(x_1,y_1)$ was already put in $U$ or $V$, then skip to the next stage. Otherwise, let $\varepsilon_1 \in E$ be sufficiently small that the open ball $B_{\varepsilon_1}(x_1,y_1)$ does not intersect symmetric ball $B_{\varepsilon_1}(y_1,x_1)$, nor does either ball contain any points that were already put in $U$ or $V$. This is always possible since $\varepsilon_0$ is not a possible value of $d$, which means that $$\min\{d((x_0,y_0),(x_1,y_1)), d((y_0,x_0),(x_1,y_1))\} > \varepsilon_0.$$ So making sure that $$\varepsilon_1 < \min\{d((x_0,y_0),(x_1,y_1)),d((y_0,x_0),(x_1,y_1))\} - \varepsilon_0$$ will do for the second requrement. Finally, put all points of $B_{\varepsilon_1}(x_1,y_1)$ in $U$ and all points of $B_{\varepsilon_1}(y_1,x_1)$ in $V$.

Continue in the same manner for all the remaining points $(x_2,y_2),(x_3,y_3),\ldots$  Since all points of $X^2$ minus the diagonal will eventually be considered, in the end we will have a partition of $X^2$ minus the diagonal into two disjoint open sets $U$ and $V$ such that $(x,y) \in U$ iff $(y,x) \in V$.