# Why doesn't separate continuity imply continuity?

Suppose $$f: U \rightarrow R$$ for some open subset $$U$$ of $$R^2$$ is continuous in each variable ie. $$f(- , y)$$ continuous for each fixed y, and $$f(x , -)$$ continuous for each fixed x.

I know the counterexample that $$f = \frac{xy}{x^2 +y^2}$$ for $$(x,y) \neq (0,0)$$ , $$f = 0$$ for $$(x,y) = (0,0)$$ is separately continuous but not continuous at the origin.

Where does the following proof that it should be continuous fail?

Suppose we try to show continuity at $$(x_1 , y_1)$$. Then for any $$(x_2, y_2)$$ in $$U$$, $$|f(x_1 , y_1) -f(x_2 , y_2)| \leq |f(x_1 , y_1) -f(x_2 , y_1)| + |f(x_2 , y_1) -f(x_2 , y_2)|$$ by the triangle inequality.

Fix $$\epsilon > 0$$.

Then $$\exists a>0$$ s.t $$|x_1 - x_2| < a \implies |f(x_1 , y_1) -f(x_2 , y_1)| < \epsilon$$.

Similarly $$\exists b>0$$ s.t $$|y_1 - y_2| < b \implies |f(x_2 , y_1) -f(x_2 , y_2)| < \epsilon$$.

Let $$\delta = min(a,b)$$, then for $$|(x_1,y_1) - (x_2,y_2)|<\delta$$, we have $$|f(x_1,y_1) - f(x_2,y_2)|< 2\epsilon$$. Done.

Is it because whilst it may work for that particular choice of $$(x_2, y_2)$$, there may be another choice, also within distance $$\delta$$ of $$(x_1, y_1)$$, such that $$|f(x_1,y_1) - f(x_2,y_2)| > \epsilon$$? If I add the condition that $$f$$ is Lipschitz in $$y$$, say, with Lipschitz constant independent of $$y$$, how is this sufficient for continuity?

That proof fails because that $$b$$ depends on $$x_2$$. It would wourk if you could choose $$a$$ and $$b$$ depending on $$x_1$$ and on $$y_1$$ alone.