Countable Union of Overlapping Connected Sets is Connected

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $\sequence {A_n}$ be a sequence of connected subsets of $T$ with:


 * $A_i \cap A_{i + 1} \ne \O$

for each $i$.

Then:


 * $\ds \bigcup_{i \mathop = 1}^\infty A_i$

is connected.

Proof
Write:


 * $\ds A = \bigcup_{i \mathop = 1}^\infty A_i$

Let $f : A \to \set {0, 1}$ be a continuous function, where $\set {0, 1}$ is given the discrete topology.

We show that $f$ is constant, so $A$ is connected.

From Restriction of Continuous Mapping is Continuous: Topological Spaces, we have:


 * $f {\restriction_{A_i}}$ is continuous on $A_i$.

Since $A_1$ is connected, we have that:


 * $f {\restriction_{A_1}}$ is constant.

That is:


 * $\map f {A_1} = \set \delta$

for some $\delta \in \set {0, 1}$.

We claim that:


 * $\map f {A_i} = \set \delta$

for each $i \in \N$.

If we can show this to be true, we have:

So $f$ is necessarily constant, meaning $A$ is connected.

We proceed by induction.

For all $i \in \N$, let $\map P i$ be the proposition:


 * $\map f {A_i} = \set \delta$

Basis for the Induction
We have shown that:


 * $\map f {A_1} = \set \delta$

So $\map P 1$ is seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P i$ is true, where $i \ge 1$, then it logically follows that $\map P {i + 1}$ is true.

So we need to show that:


 * $\map f {A_i} = \set \delta$

for some $i \in \N$, implies that:
 * $\map f {A_{i + 1} } = \set \delta$

Induction Step
Note that $A_{i + 1}$ is connected, so:


 * $f {\restriction_{A_{i + 1} }}$ is constant.

So:


 * $\map f {A_{i + 1} } = \set \phi$

Since $A_i \cap A_{i + 1} \ne \O$, we have:


 * $\map f {A_i \cap A_{i + 1} } \ne \O$

From Image of Intersection under Mapping, we have:

So:


 * $\set \delta \cap \set \phi \ne \O$

This is the case $\phi = \delta$, so we obtain:


 * $\map f {A_{i + 1} } = \set \delta$

So $\map P i \implies \map P {i + 1}$ and the result follows by the Principle of Mathematical Induction.

So:


 * $\map f {A_i} = \set \delta$

for each $i \in \N$.

By our prior calculations, we are now done.