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I went to a exercise class and I got really confused. Consider the problem:

Let $\lbrace A_{n} \rbrace$ be a sequence of connected subspaces of $X$, such that $A_{n}\cap A_{n+1}\neq \emptyset$ for all $n$. Show that $\cup A_{n}$ is connected.

My TA started by showing that for all $n\in \mathbb{N}$ $B_{n} = \bigcup_{j=1}^{n} A_{j}$ is connected using induction. I understand this part. But I would finish the proof here, because if you now this holds for all natural numbers $n$ why can't you just deduce that it holds for $\bigcup_{n=1}^{\infty} A_{n}$? My TA didn't finished the proof here, but used the result to finish his proof in another way.

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up vote 20 down vote accepted

There is a difference between knowing that each finite union is connected and knowing that the infinite union is connected. Let me give you an example, using a different property, that shows the difference very clearly.

Suppose that $A_n=\{n\}$ for each $n\in\Bbb Z^+$. Then it’s certainly true that $\bigcup_{k=1}^nA_k$ is finite for each $n\in\Bbb Z^+$, but it’s certainly not true that $\bigcup_{k\ge 1}A_k$ is finite. If you prefer a more topological example, think of each $A_n$ as a subset of $\Bbb R$. Then $\bigcup_{k=1}^nA_k$ is compact for each $n\in\Bbb Z^+$, but $\bigcup_{k\ge 1}A_k$ is not compact.

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Thanks! I see the problem. Using an induction proof I can only conclude that given a natural number the statement holds, but it requires more details if I want to show that it holds when we let $n$ tend to infinity. – guestfrom Jan 8 '13 at 16:41
@guestfrom: You’re welcome! – Brian M. Scott Jan 8 '13 at 16:46

Lemma: $\infty$ is finite.

Proof: $1$ is finite. If $n$ is finite, then $n+1$ is finite. Thus by induction all natural numbers are finite. Thus $\infty$ is finite.

This fails because $\infty$ is not a natural number. There is no reason that something holding true for any natural number means it should hold true for $\infty$.

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