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Suppose the restrictions of a function $f$ to sets $E_1, ..., E_n$ are continuous. Then is the restriction of $f$ to $\cup_{i=1}^{N} E_i$ continuous? What about to $\cup_{i=1}^{\infty} E_i$?

It seems like common sense but I'm not sure if it's true, especially for the infinite case.

Thanks!

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    A [similar post](http://math.stackexchange.com/questions/106896/continuity-and-partition-of-domain)2012-02-09

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Let $f:\mathbb R \to \mathbb R$ be the indicator function of $\mathbb Q$, 1 on the rationals and 0 on the irrationals. Its restrictions to $\mathbb Q$ and $\mathbb R - \mathbb Q$ are continuous (they are constant functions) but $f$ is not continuous on $\mathbb R = \mathbb Q \cup (\mathbb R - \mathbb Q)$.