some confusion in cofinite topology..

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Let $\{A_\alpha\}$ be a collection of subsets of some topological space $X$ s.t. $X=\cup_{\alpha}A_\alpha.$ Let $f:X\to Y$, where $Y$ is some other topological space. Suppose that $f|A_\alpha$ is continuous for each $\alpha$. Find an example where the collection $\{A_\alpha\}$ is countable and each $A_\alpha$ is closed, but $f$ is not continuous. [Munkres 2/e, 18.9(b)]

i found the answer but i did n't understand the red line .....enter image description here

Can any body help me understanding the red lines?,,,as im not getting and understand in my head.....

thanks in advance

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The only integer contained in $(-1/2, 1/2)$ is $0$, so we have that $f^{-1}(-1/2, 1/2) = \{0\}$, since $f$ is the identity map (or inclusion map). Now $(-1/2, 1/2)$ is open in $\mathbb{R}$ with the standard topology, so if $f$ is continuous, its preimage $\{0\}$ should be open in $\mathbb{Z}$ with the cofinite topology. The open sets in the cofinite topology are by definition the empty set and any subset whose complement is finite. Thus $\{0\}$ is not open in the cofinite topology on $\mathbb{Z}$ since $\mathbb{Z}\setminus \{0\}$ is not finite. Thus $f$ is not continuous.