# Real Analysis – Prove that a function with disjointed open intervals is continuous

It works because continuity is a local property. To repair $$x_0 in (0,1) cup (1,2)$$; I will handle the case with $$x_0 in (0,1)$$ since a symmetric argument applies if $$x_0 in ( 1,2)$$. Now we know that $$f$$ is continuous on $$(0.1)$$. So, given $$epsilon> 0$$, we can find $$delta> 0$$ such as
$$| f (x) – f (x_0) | < epsilon$$
does not matter when $$x in (0,1)$$ satisfied $$| x-x_0 | < delta$$. Because $$(0.1)$$ is open, there is $$delta ^ premium> 0$$ so small that
$$(x_0- delta ^ prime, x_0 + delta ^ prime) subseteq (0,1)$$
and especially, $$(x_0- delta ^ prime, x_0 + delta ^ prime) cap (1,2) = do not change anything$$. It gives us everything we need.

To choose $$0 < delta ^ { prime prime} < min ( delta, delta ^ prime)$$ and let $$x in (0,1) sqcup (1,2)$$ to be such that $$| x-x_0 | < delta ^ { prime prime}$$. By our choice of $$delta ^ premium$$, we must have $$x, x_0 in (0,1)$$ so that
$$| h (x) -h (x_0) | = | f (x) -f (x_0) | < epsilon$$
where we used that $$| x-x_0 | < delta ^ { prime prime} < delta$$ in this last inequality.

Essentially, we chose $$delta ^ { premium prime}> 0$$ so small that everything $$x$$ with a $$delta ^ { premium prime}$$-distance $$x_0$$ live in the open decor $$(0.1)$$. Therefore, $$h$$ will only be determined by the continuous function $$f$$ on the open interval $$(x_0- delta ^ { premium prime}, x_0 + delta ^ { premium prime})$$. As mentioned above, the same argument applies if $$x_0 in ( 1,2)$$ instead of.