From the pasting lemma we have that if ## X = A \cup B ## and ##f: A \rightarrow Y ## and ## g: A \rightarrow Y## are continuous functions that coincide on ## A \cup B ##, they combine to give a continuous function ## h: X \rightarrow Y ## s.t. ## h(x) = f(x) ## for ## x \in A ## and ## h(x) = f(g) ## for ## x \in B ##, so long as both ## A ## and ## B## are both open or closed.(adsbygoogle = window.adsbygoogle || []).push({});

But what if ##A## is open and ##B## is closed? This should not work. My reasoning is that, if it worked, then the pasting lemma would already say so! But this obviously does not constitute a proof or counterexample. I think it shouldn't be difficult, but I am missing something. The "obvious" choice was to take ##A## and ##B## as complements, but I am not sure how to show a lack of continuity. The intersection is empty, so the functions agree on that. Another example?

I feel like this should be very easy or obvious. A nudge in the right direction would be appreciated.

-Dave K

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# A A pasting lemma counterexample (of sorts)

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