Return to Answer

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$---in particular, it is impossible when $I$ is nonempty and well-ordered---since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered (or nonempty and well-ordered, depending on whether you consider $\emptyset$ to be connected).

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a nonempty well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. Lastly, if $I$ is empty, then the hypothesis of 2 is vacuous, but the union set is empty, which is technically disconnected. QED

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$ — in particular, it is impossible when $I$ is nonempty and well-ordered — since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered (or nonempty and well-ordered, depending on whether you consider $\emptyset$ to be connected).

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a nonempty well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. Lastly, if $I$ is empty, then the hypothesis of 2 is vacuous, but the union set is empty, which is technically disconnected. QED

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$---in particular, it is impossible when $I$ is nonempty and well-ordered---since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered.

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a nonempty well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. QED

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$---in particular, it is impossible when $I$ is nonempty and well-ordered---since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered (or nonempty and well-ordered, depending on whether you consider $\emptyset$ to be connected).

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a nonempty well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. Lastly, if $I$ is empty, then the hypothesis of 2 is vacuous, but the union set is empty, which is technically disconnected. QED

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$---in particular, it is impossible when $I$ is nonempty and well-ordered---since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered.

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. Suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. QED

I assume that you intend $I$ is linearly ordered by $<$. The hypothesis is impossible when $I$ has a least element $\beta$---in particular, it is impossible when $I$ is nonempty and well-ordered---since if $\beta$ is the least element, then $\bigcup_{\alpha<\beta}A_\alpha$ is empty itself. But if we insist on that requirement only when $\beta$ is not the least element of $I$, then your conclusion holds if and only if $I$ is well-ordered.

Theorem. The following are equivalent for any linearly ordered set $\langle I,<\rangle$.

1. $\langle I,<\rangle$ is a nonempty well-order.

2. Whenever $X$ is a topological space and $A_\alpha\subset X$ is a connected subspace for every $\alpha\in I$ and whenever $\beta\in I$ is not least, then $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, then $\bigcup_{\alpha\in I}A_\alpha$ is connected.

Proof. ($1\to 2$) Assume $I$ is nonempty and well-ordered by $<$. We prove that $\bigcup_{\alpha\in I}A_\alpha$ is connected by transfinite induction on the order type of $I$. Suppose that $U\sqcup V$ is a nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$. Let $\beta$ be least such that $\bigcup_{\alpha\leq\beta}A_\alpha$ has points from both $U$ and $V$. But this union is equal to $\bigcup_{\alpha<\beta}A_\alpha\cup A_\beta$, which by induction is the union of two overlapping connected subspaces, and hence is connected. So it cannot have points from both sides of the separation, a contradiction. So there is no nontrivial open separation of $\bigcup_{\alpha\in I}A_\alpha$, and so it is connected.

($2\to 1$) We prove the contrapositive. For the main case, suppose that $I$ is not a well-order. Thus, it has a strictly descending sequence $\alpha_0>\alpha_1>\alpha_2>\cdots$ and so on. If $\alpha$ is below all the $\alpha_n$, then let $A_\alpha=\{0\}$ have just the point $0$. Otherwise, $\alpha_n\leq \alpha$ for some smallest $n$, and in this case, we let $A_\alpha$ be $\{0\}$, if $n$ is even, and otherwise $\{1\}$. So each $A_\alpha$ is connected, and furthermore, for every $\beta\in I$ (except the least element, if there is one), we have $\bigcup_{\alpha<\beta}A_\alpha\cap A_\beta\neq\emptyset$, since below the descending sequence, both of these sets are just $\{0\}$, and above any point in the descending sequence, the union set is $\{0,1\}$, which meets $A_\beta$. Meanwhile, the union $\bigcup_{\alpha\in I}A_\alpha$ is $\{0,1\}$, which is not connected, by design. QED