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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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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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