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.

  • 1
    $\begingroup$ Sorry, in the title the requirement of open subsets was stated, it got lost in the body of the question. $\endgroup$
    – Jochen
    May 23, 2021 at 11:39

1 Answer 1


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.)

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

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