Consider the set $\{A_i\}_{i = 1}^{\infty}$ be a set of topological spaces which are connected. If $A_i \cap A_{i+1} \neq \emptyset$ for each $i \geq 1$ then it's an easy exercise to show that $\bigcup_{i=1}^{n} A_i$ is connected (1).

Since we usually do not deal with countable collection of spaces, I would like to extend this to an arbitrary indexing set $J$. The two almost obvious conditions that can be given to $A_{\alpha}'s$ are below

  1. $\{A_{\alpha}\}_{\alpha \in J}$ where $A_{\alpha}$'s are pairwise not disjoint
  2. For $\{A_{\alpha}\}_{\alpha \in J}$, fix an $\alpha_0$ such that $A_{\alpha_{0}} \cap A_{\alpha} \neq \emptyset$ for all $\alpha \in J$

(Case 2 is also present in Case1, but I have written it here just for the sake of completeness)

The way I want to extend the initial statement is not like the above cases. Giving a more rough geometric picture, the condition $A_i \cap A_{i+1} \neq \emptyset$ gives a more "chain" like realization for the space $\bigcup_{i}A_i$. My idea is to extend that geometric picture. One thing to realize is that such a construction needs an order relation on $J$. So I will be assuming that $J$ has an ordered set with the order $\leq$

One level of generalization could be to consider an ordered set $J$ such that for each $a_i$ there exists elements $a_{i-1}$ and $a_{i+1}$ such that $a_{i-1} \leq a_i \leq a_{i+1}$. For such an indexing set $J$, the set of spaces $\{A_a\}_{a \in J}$ has the property that $\bigcup_{a \in J} A_a$ is connected (Proof: To show this consider a continuous function $f : \bigcup_{a \in J} A_a \to \{0,1\}$ , since the restriction of a continuous function is also continuous, specifically restricting it to each $A_a$ is also continuous. But $A_a$ for all $a\in J$ is connected and hence the function restricted to $A_a$ is a constant function. Also $A_{a_i} \cap A_{a_{i+1}} \neq \phi$ for all $a_i \in J$.Hence $f$ is constant and the space is connected). But such an indexing set is bijective to set of integers $\Bbb{Z}$ hence is still countable.

In order to further generalize it to an arbitrary indexing set J which doesn't have the successor/predecessor property, this is how I proceeded:

Claim: Given $\{A_a\}_{a \in J}$ a collection of topological spaces which are connected. If the following three conditions are satisfied, I claim that $\bigcup_{a \in j} A_a$ is connected.

  1. For each $a \in J$( where $a$ is not the smallest/greatest element of the set $J$), $\exists$ $\beta,\delta$ such that $\beta \leq a \leq \delta$ such that for each $\gamma \in [\beta,\delta]$(a notation to represent elements lying between $\beta$ and $\delta$) $A_a \cap A_{\gamma} \neq \emptyset$
  2. If $a$ is the least element of $J$, $\exists \beta \geq a$ such that $A_a \cap A_{\gamma} \neq \emptyset$ for $\gamma \in [a, \beta]$
  3. Similarly if $a$ is the greatest element, $\exists \beta \leq a$ such that $A_a \cap A_{\gamma} \neq \emptyset$ for $\gamma \in [\beta , a]$

Questions: 1) Given the way I have proceeded to define the intersections of the connected sets, I have been sadly been unable to prove that $\bigcup{a \in J} A_a$ is connected. How should I go about in proving such a thing given my "elaborate" conditions

2) Is the above method a right/sensible/natural way to generalize the initial question (1) keeping the chain picture in mind? . I didn't look for a counterexample but is there a counterexample to (1) for an arbitrary indexing set $J$ which has an order on it?

Thanks in advance for going through all that.

  • $\begingroup$ You're intermediate conclusion that if you have an ordered set $J$ (totally oredered or partially ordered?) so that for any $a\in J$ $A_a$ is connected to an $A_b$ and $A_c$ with $b≤a≤c$ is false. Consider for example $J=\Bbb Z_1 \cup \Bbb Z_2$ two copies of $\Bbb Z$ where every element in the second is larger than every element in the first copy. Then let $A_n= \{0\}$ on the first copy and $A_n=\{1\}$ for hte second copy. $\endgroup$
    – s.harp
    Sep 19, 2019 at 18:23
  • 1
    $\begingroup$ Your main claim is also incorrect, consider $J=\Bbb R-\{0\}$ and $A_a=\{0\}$ for negative $a$ and $A_a=\{1\}$ for positive $A$. Note that for any $a\in J$ you have some very small $\epsilon$ with $[a-\epsilon, a+\epsilon]$ not crossing over $0$. $\endgroup$
    – s.harp
    Sep 19, 2019 at 18:26
  • $\begingroup$ Give $J$ the structure of a graph: points $a,b$ are connected if $A_a\cap A_b\neq\emptyset$. If $J$ is connected with this graph structure and $A_a$ are also all connected then you will get the conclusion that $\bigcup_{a\in J} A_a$ is connected. How can you guarantee in simple or natural way that $J$ is connected? You could for example say that for any two $a, b\in J$ there is a $c\in J$ with $A_a\cap A_c \neq\emptyset\neq A_b\cap A_c$. Connecting this to partial orders you may say: $J$ is a directed set and for any two $a,b$ there is a $c≥a, c≥b$ so that $A_a$ and $A_b$ connect to $A_c$. $\endgroup$
    – s.harp
    Sep 19, 2019 at 18:34
  • $\begingroup$ @s.harp Could you elaborate on the example you gave for the intermediate conclusion? I don't seem to understand your example on how $A_n$ given that A_n are connected is taking two different values? $\endgroup$ Sep 19, 2019 at 19:09
  • 1
    $\begingroup$ Your condition is that every element in $J$ has a predecessor and a successor. $J=\Bbb Z_1\cup \Bbb Z_2$ with the usual order on the copies of $\Bbb Z$ together with $\Bbb Z_1≤ \Bbb Z_2$ has this property. Now if $n\in\Bbb Z_1$ let $A_n = \{0\}$, if $n\in\Bbb Z_2$ let $A_n=\{1\}$. Now for any $n\in\Bbb Z_1\cup\Bbb Z_2$ its predecessor and successor are in the same copy of $\Bbb Z$, hence $A_{n-1}=A_n=A_{n+1}$. But clearly the union of everything is not connected. $\endgroup$
    – s.harp
    Sep 19, 2019 at 20:10

2 Answers 2


Your claim is not true, mostly because linearly ordered index sets can be wilder you thought:

Let $I = \mathbb{Z} \times \{0,1\}$ in the reverse lexicographic order (anything with 0 as the second coordinate comes before anything with 1 as the second coordinate, and when the second coordinate is equal, the first one decides the order), and $X=\Bbb R$ in the standard topology and $A_i=[0,1]$ if $i$ is of the form $(n,0)$ and $[2,3]$ otherwise. $I$ has no minimum or maximum so those clauses are irrelevant and for any $i=(n,j)$ we can take as the required interval $[(n-1,j),(n+1,j)]$, e.g. So the setup obeys all your requirements yet the union is not connected.


As remarked in the comments and the answer by Henno Brandsma, the main claim is false.

In order to recover an analogous statement we shift perspective a bit. Give the indexing set $J$ the structure of an undirected graph, were two points $a,b\in J$ are connected if $A_a\cap A_b\neq\emptyset$. Clearly $\bigcup_{a\in J}A_a$ is connected if the graph on $J$ is connected. The goal is to find natural or simple conditions that guarantee that this graph $J$ is connected.

First we must remark that while asking for $J$ to be connected is the right condition, it is not an iff condition, only: $\left(\bigcup_{a\in J}A_a\text{ is connected }\impliedby\text{ $J$ is connected}\right)$ is true. For example if $X$ is a connected topological space, take $J=X$ and $A_x=\{x\}$ for all $x\in J$. Clearly $\bigcup_{x\in J}A_x=X$ is connected, but the graph $J$ is discrete as no two $A$ interesect. But as this example makes clear there can be no sensible condition on the intersection structure to recover an $\iff$.

When can you guarantee that the graph on $J$ is connected? For $J=\Bbb Z$ or $J=\Bbb N$ you have the inherent notion of successor, asking that every vertex of $J$ is connected to its successor results in a connected graph, giving the examples you started out with.

Another example of a common index set is if $J$ has the structure of a directed set. An example of a directed set is the power set $\mathfrak{P}(X)$ for any set $X$. Here a set $a≤b$ if $a\subseteq b$, but not any two sets are in relation to one-another by $≤$. However for $a,b\in \mathfrak{P}(X)$ you have that $a≤a\cup b$ and $b≤a\cup b$, that is there is an element greater than both $a$ and $b$. Infact $\mathfrak P(X)$ is a join semilattice, in that a least upper bound of $a,b$ exists (its actually a lattice since you also have a meet $\cap$, but thats not relevant here).

For join semilattices the following condition is natural: For any $a, b$ the sets $A_a, A_b$ are both connected to $A_{a\cup b}$ where $a\cup b$ is the join of $a,b$.


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