1

Given an open cover $\{G_t\}_{t \in T}$ of connected space $X$ and two points $a, b \in X$ show that there exists finite sequence of indices $t_1, t_2, \dots, t_n$ (a snake) such that $G_{t_i} \cap G_{t_{i+1}} \neq \varnothing $ for $i = 1, 2, \dots, n-1$ and $a \in G_{t_0}$, $b \in G_{t_n}$.

Hint: show that the set of points $x \in X$ that can be connected with $a$ using snakes is open and closed in $X$.

Is this even true? I thought that the long line could be a counter-example, which is path-connected (and therefore connected) and setting $a = (0, 0)$, $b = (1, 0)$, where long line is $\omega_1 \times [0,1)$.

  • What is $(1,0)\in\omega_1\times[0,1)$? Which element of $\omega_1$ does $1$ denote? – Stefan Hamcke Jun 05 '15 at 18:32
  • @StefanHamcke: Sorry for not being formal. We assume generalized continuum hypothesis and therefore we identify $\omega_1$ with $\mathbb R$. Using Axiom of Choice we can well-order $\mathbb R$, so I think that my argument fails. So maybe the statement is true... How to prove it? –  Jun 05 '15 at 19:50
  • 2
    Just use the hint :-) – Stefan Hamcke Jun 05 '15 at 19:58
  • 1
    I gave a complete answer here: http://math.stackexchange.com/a/44938/4280 – Henno Brandsma Jun 06 '15 at 07:47

2 Answers2

4

Let $O = \{x \in X: x \text{ can be connected by a snake to } a\}$, as the hint says.

Clearly $O \neq \emptyset$: let $G_{t_1}$ be a member of the cover that contains $a$, then $(t_1)$ is a "snake" of length $n=1$ (the intersection condition is thus void) that connects $a$ to itself (as $a \in G_{t_1}$).

Suppose $x \in O$, so we have some finite sequence $t_1,\ldots,t_n$ that is a snake from $a$ to $x$, so $x \in G_{t_n}$ and $a \in G_{t_1}$ plus the intersection condition. Note that for all $y \in G_{t_n}$ the same sequence is a snake from $a$ to $y$ as well; the only thing we need to verify is the trivial $y \in G_{t_n}$, and so such a $y$ is in $O$ as well. So $x \in G_{t_n} \subseteq O$, and this show that any $x \in O$ is an interior point of $O$, so $O$ is an open set.

To see that $O$ is closed, let $y \in X \setminus O$ be arbitrary. Now pick $u \in T$ such that $y \in G_u$ (using that we have a cover of $X$). Suppose $G_u$ intersects $O$, say in $x$. Then for $x$ there is some snake $t_1,\ldots,t_n$ from $a$ to $x$, but then $t_1,\ldots,t_n,u$ is a chain from $a$ to $y$: $y \in G_u$ and the last intersection (the only new one in the snake, not covered by the condition in the old one) $G_{t_n} \cap G_u$ is also non-empty as witnessed by $x$. But this contradicts that $y \in X \setminus O$. So this shows that $y \in G_u \subseteq X \setminus O$ and so $y$ is an interior point of the complement of $O$, making the latter set open, and so $O$ closed.

Now connectedness of $X$ tells us that $O = X$, as the only non-empty closed-and-open set of $X$ is $X$ in that case. In particular $b \in O$, which is what you want.

Henno Brandsma
  • 250,824
  • Is the converse true? –  Apr 25 '17 at 12:50
  • Sure. Suppose $X$ has the snake property for all open covers. Then $X$ is connected. @416333 – Henno Brandsma Apr 25 '17 at 12:52
  • I don't see it. What if the elements of the cover were "disconnected"? –  Apr 25 '17 at 13:24
  • Suppose $X$ were disconnected. Then $X =U\cup V$ where both sets are open disjoint and non-empty. Then the snake property fails for this two set cover and two points in different members. @416333 so connected implies snake property and not connected implies not snake property. Hence equivalence – Henno Brandsma Apr 25 '17 at 13:39
  • Wow! That was simple. Thank you very much! –  Apr 25 '17 at 13:55
2

An alternative way of thinking about this problem: For $x, y \in X$, write $x \sim y$ if there exists a snake connecting $x$ to $y$. Claim: this relation on $X$ defines an equivalence relation, and the equivalence classes are open.

If the claim is true, then the existence of snakes for a connected space is an easy corollary: $X$ is the disjoint union of open sets given by the equivalence classes $$[x] := \{y \in X \mid x \sim y\}$$ for $x \in X$. If $\{U_i\}_{i \in I}$ denotes the family of equivalence classs, then as already stated, we have $$X = \bigsqcup_{i \in I} U_i$$ and for a fixed $x \in X$, $[x]$ is open, $[x]=U_j$ for some $j \in I$, and $$X \setminus [x] = \bigsqcup_{\substack{i \in I \\ i \neq j}} U_i = \bigcup_{\substack{i \in I \\ i \neq j}} U_i$$ is open, so $[x]$ is both open and closed. If $X$ is connected, this implies $X = [x]$, so every $y \in X$ can be connected to $x$ by a snake.

To show that $\sim$ is indeed an equivalence relation:

  • Reflexive: For $x \in X$, pick $t \in T$ so that $x \in G_t$. Then the trivial sequence $t$ connects $x$ to itself, so $x \sim x$.
  • Symmetric: For $x, y \in X$ with $x \sim y$, let $t_1, \ldots, t_n$ be a snake connecting $x$ to $y$. Then $t_n, \ldots, t_1$ is a snake connecting $y$ to $x$, so $y \sim x$.
  • Transitive: Let $x, y, z \in X$ with $x \sim y$ and $y \sim z$. Let $t_1, \ldots, t_n$ be a snake connecting $x$ to $y$, and let $s_1, \ldots, s_k$ be a snake connecting $y$ to $z$. Then $y \in G_{t_n} \cap G_{s_1}$, so in particular, $G_{t_n} \cap G_{s_1}$ is nonempty, and $t_1, \ldots, t_n, s_1, \ldots, s_k$ is a snake connecting $x$ to $z$, thus $x \sim z$.

And finally, if $x \in X$, then the claim is that the equivalence class of $x$ under the equivalence relation $\sim$, $[x]$, is open. This is clear: If $y \in [x]$, then there is a snake $t_1, \ldots, t_n$ connecting $x$ to $y$, and we therefore have $y \in G_{t_n} \subset [x]$.

Dustan Levenstein
  • 13,219