# Thread: proving continuity of g(x, y) = x

1. ## proving continuity of g(x, y) = x

g(x, y) = x

I am trying to prove that g is continuous on $R^2$ using the epsilon-delta definition of continuity.

I gave it a try but it doesn't look right:

Given $c \in R$ and $\epsilon > 0$, set $\delta = \epsilon$

Then whenever $\| x-c \| < \delta$, it follows that $\| g(x, y) - g(c, y) \| < \epsilon$

Thanks for any help in advance.

2. $|g(x,y)-g(x_0,y_0)|<\epsilon$ tells you that the distance between x and x_0 is less than epsilon (i.e. $|x-x_0|<\epsilon$).

$||(x,y)-(x_0,y_0)||<\delta$ tells you that $|x-x_0|\leq\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$.

Do you see why you get continuity by picking delta as you did?

3. Originally Posted by Focus
$|g(x,y)-g(x_0,y_0)|<\epsilon$ tells you that the distance between x and x_0 is less than epsilon (i.e. $|x-x_0|<\epsilon$).

$||(x,y)-(x_0,y_0)||<\delta$ tells you that $|x-x_0|\leq\sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$.

Do you see why you get continuity by picking delta as you did?
Oh, okay. I think I get it now.

If $\delta = \epsilon$,
$|x-x_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ would imply
$|x-x_0| < \epsilon$ which would imply
$|g(x,y)-g(x_0,y_0)| < \epsilon$

Edit: also I'd be given $(x_0, y_0) \in R^2$, not just $c \in R$ since g is in R^2

4. Originally Posted by DPMachine
Oh, okay. I think I get it now.

If $\delta = \epsilon$,
$|x-x_0| \leq \sqrt{(x-x_0)^2+(y-y_0)^2}<\delta$ would imply
$|x-x_0| < \epsilon$ which would imply
$|g(x,y)-g(x_0,y_0)| < \epsilon$

Edit: also I'd be given $(x_0, y_0) \in R^2$, not just $c \in R$ since g is in R^2
Looks good to me. Good luck with your studies.