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$X = \mathbb{R}$, $\mathcal{T} =$ the collection of all subset $U$ of $\mathbb{R}$ such that $U = \emptyset$ or $\mathbb{R} - U$ is finite.

Then, $(X,\mathcal{T})$ is connected.

My thoughts: assume not connected, there is $A, B$ that forms separation of $X$. and $A \cap B = \emptyset$. We have $X - A \cap B = X \implies (X-A)\cup(X-B) = X$ where $X-A$ and $X-B$ finite, but $\mathbb{R}$ is infinite set, cannot be the union of two finite sets. Contradiction

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  • $\begingroup$ The idea is correct, but there is what seems to be a typo: $$(X-A)\cup(X-B)=X$$ $\endgroup$ – ajotatxe Jul 27 '15 at 18:03
  • $\begingroup$ @ajotatxe, thx, have it fixed $\endgroup$ – ElleryL Jul 27 '15 at 18:05
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Your proof is correct, but you’ve not expressed it very well. Here’s a clearer version:

If $X$ is not connected, there are disjoint non-empty open sets $A$ and $B$ such that $X=A\cup B$. Since $A$ and $B$ are non-empty, $X\setminus A$ and $X\setminus B$ are finite. But $X\setminus A=B$, and $X\setminus B=A$, so $A$ and $B$ are both finite, and hence so is $X=A\cup B$. But this is absurd, since $X=\Bbb R$ is infinite.

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  • $\begingroup$ Thx for your answer, :) $\endgroup$ – ElleryL Jul 27 '15 at 18:09
  • $\begingroup$ @Hobbit6094: You’re welcome. $\endgroup$ – Brian M. Scott Jul 27 '15 at 18:12

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