# Can the increasing union of open disconnected sets be connected?

Let $$A_n$$ be an increasing sequence of open subsets of the sphere $$S^2$$ such that the union is connected (hence also path-connected). Does this imply that at least one $$A_n$$ is connected?

The strategy should be that the image of a continuous path between two points is compact and hence contained in some $$A_n$$. But this is not exactly what one needs. Trying to argue by contradiction, one could fix $$x_0\in A_0$$ and assume that there are $$x_n\in A_n$$ which cannot be connected to $$x_0$$ by a path in $$A_n$$. Using the compactness of $$S^2$$ one gets a limit $$z$$ of some subsequence. If $$z$$ were in some $$A_k$$, one could find a path-connected neighbourhood $$U$$ of $$z$$ contained in $$A_k$$. Morevoer, a path from $$x_0$$ to $$z$$ is contained in some $$A_m$$ for $$m\ge k$$ such that $$x_m\in U$$. Then one can concatenate the path from $$x_0$$ to $$z$$ with a path in $$U\subseteq A_m$$ from $$z$$ to $$x_m$$ to get a path in $$A_m$$ from $$x_0$$ to $$x_m$$. However, the limit point $$z$$ need not be in the union, so that the argument fails.

I could imagine a homology argument like $$H_0(\bigcup A_n)$$ is the colimit in the category of abelian groups of $$H_0(A_n)$$ and if the former in $$\mathbb Z$$ also one of the $$H_0(A_n)$$ has to be $$\mathbb Z$$. Without further details this is just hand waving and not at all a proof. Anyway, the equivalence between path-connectedness and $$H_0(A)=\mathbb Z$$ is not very deep, and one should be able to transform a homological proof into an elementary one.

On the other hand, homology theory might suggest that some hypothesis -- perhaps $$H_1(A_n)=0$$? -- might be missing. If this is so, then a translation into an elementary proof would be tricky.

• Sorry, in the title the requirement of open subsets was stated, it got lost in the body of the question. May 23, 2021 at 11:39

For simplicity, here's a counterexample in $$\mathbb{R}$$ instead of $$S^2$$:

$$A_n=(0,n)\cup (n+{1\over 3}, n+{2\over 3}).$$ Each $$A_n$$ is open and disconnected and contained in $$A_{n+1}$$, but $$\bigcup A_n=(0,\infty)$$ is connected. Intuitively, the $$A_n$$s are not disconnected in the same way.

To move this from $$\mathbb{R}$$ to $$S^2$$, simply work with open discs or squares instead of open intervals. For instance, let $$h:\mathbb{R}^2\rightarrow S^2$$ be the usual embedding missing a single point (or just any embedding at all), let $$U_n=(0,n)^2\cup (n+{1\over 3}, n+{2\over 3})^2$$, and let $$A_n=h[U_n]$$.

That said, here's a simple positive result:

Fix a topological space $$\mathcal{X}$$, and suppose we have sets $$A_n,U_n,V_n$$ (for $$n\in\mathbb{N}$$) such that:

• The sequence $$(A_n)_{n\in\mathbb{N}}$$ is increasing.

• Each $$U_n$$ and $$V_n$$ is nonempty and open, and $$U_n\cap V_n=\emptyset$$.

• $$U_n\cap A_n\not=\emptyset$$ and $$V_n\cap A_n\not=\emptyset$$.

• $$U_n\cup V_n\supseteq A_n$$.

• $$U_n\subseteq U_{n+1}$$ and $$V_n\subseteq V_{n+1}$$.

Then $$\bigcup_{n\in\mathbb{N}}A_n$$ is disconnected. (Proof: consider $$\bigcup_{n\in\mathbb{N}}U_n$$ and $$\bigcup_{n\in\mathbb{N}}V_n$$.) Note that this does not require the $$A_n$$s to be open.

So in fact the issue above is the only possible obstacle: if we have a "uniform method" for disconnecting the $$A_n$$s, then their union is disconnected.

(What if we just demand $$U_n\cap A_n\subseteq U_{n+1}\cap A_{n+1}$$ and $$V_n\cap A_n\subseteq V_{n+1}\cap A_{n+1}$$ instead of $$U_n\subseteq U_{n+1}$$ and $$V_n\subseteq V_{n+1}$$? Well, if we require the $$A_n$$s to be open we're fine: just replace $$U_n,V_n$$ by $$U_n\cap A_n, V_n\cap A_n$$ and apply the result above. If we don't add this requirement, though, we need not be able to "extend disconnections" in the obvious way! For instance, working in $$\mathbb{R}$$ let $$A_n=\{{z\over 3^n}: z\in\mathbb{Z}\}$$, $$U_n=\{({2z\over 3^n}-{1\over 10^n}, {2z\over 3^n}+{1\over 10^n}): z\in\mathbb{Z}\}$$, and $$V_n=\{({2z+1\over 3^n}-{1\over 10^n}, {2z+1\over 3^n}+{1\over 10^n}): z\in\mathbb{Z}\}$$. So here's an interesting follow-up question: under the weaker "increasing modulo $$A_n$$s" assumption, and not assuming that the $$A_n$$s are open, can we get a connected union? Note that the example two sentences prior does not give a connected union.)

• But the sequence is not increasing. May 23, 2021 at 11:40
• @Jochen Yes it is. We have $A_{n+1}\supseteq A_n$ since $(n+{1\over 3}, n+{2\over 3})\subseteq (0, n+1)= A_{n+1}$. May 23, 2021 at 11:41
• Indeed, it is increasing. May 23, 2021 at 11:42
• @Jochen So it is a counterexample to the claim in your question (once we move things from $\mathbb{R}$ to $S^2$ appropriately). May 23, 2021 at 11:43
• Yes, this is a very simple counterexample. Thanks. May 23, 2021 at 11:48