have I applied metric continuity definition correctly? if $f: R\to R$ and $g: R\to R$ are both continuous on $R$,
and if $h((x,y))=(f(x),g(gx))$, prove h is continuous on $R^2$.
this is my answer:
let $\epsilon>0$. I am going to show there exists a $\delta>0 $ s.t 
$d_{22}(h(x,y),h(x_0,y_0))<\epsilon$ whenever $d_{21}((x,y),(x_0,y_0))<\delta$
where both $d_{22}$ and $d_{11}$ must be metrics in $R^2$. right? so I will use taxicap metric. 
 from continuity of f , g and using usual metric in R we have:
$|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta_f$
$|g(y)-g(y_0)|<\epsilon$ whenever $|y-y_0|<\delta_g$
Now if I choose $\delta=\delta_f+\delta_g$ I will have:
$|f(x)-f(x_0)|+|g(y)-g(y_0)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
whenever $|x-x_0|+|y-y_0|<\delta_f+\delta_g=\delta$
I feel like something is wrong because I ve seen several example in which $\delta=min\{\delta_1.\delta_2\}$. here however I think I NEED to add them up.
other than that could anybody please give a brief explanation on how to prove a continuiuity problem using the sequence convergence definition of continuity or even open balls or open sets or... or links to examples solves using those methods? I am really looking for them but I can not find solved examples. Thanks.
 A: Even when proving continuity under the taxicab metric, this doesn't work. Given $\epsilon > 0$, the thing you are trying to prove is:

There is a $\delta > 0$ such that, for all $(x,y)$, if $d((x,y),(x_0,y_0)) < \delta$, then $d(f(x,y), f(x_0,y_0)) < \epsilon$.

There in an order to this statement. You pick a $\delta$ first, then for an otherwise arbitrary $(x,y)$ about which you only know that they satisfy the condition $d((x,y),(x_0,y_0)) < \delta$, you use that condition to prove $d(f(x,y), f(x_0,y_0)) < \epsilon$.
That is the way the proof must flow. If it doesn't flow that way, you have not proved the statement. Your "proof" does not flow this way. So let's try to fix that. 


*

*The first step is to pick a delta, and this part is good. Since $f,g$ are continuous, there do exist $\delta_f, \delta_g> 0$ such that 


For all $x, y$, if $|x - x_0| < \delta_f, |y - y_0| < \delta_g$, then $|f(x) - f(x_0)| < \frac \epsilon 2, |g(y) - g(y_0)| < \frac \epsilon 2$

Note that we do not know yet that $|x-x_0| <\delta_f$ or $|y - y_0| < \delta_g$. In fact, we haven't even introduced values of $x$ or $y$ to put in these statements. We just know that later when we do have such values $x,y$ that satisfy $|x-x_0| <\delta_f$ or $|y - y_0| < \delta_g$, then we will also know $|f(x) - f(x_0)| < \frac \epsilon 2, |g(y) - g(y_0)| < \frac \epsilon 2$.
You pick $\delta = \delta_f + \delta_g$


*

*Next, we introduce a point $(x,y) \in \Bbb R^2$. This point must be arbitrary to satisfy the "for all" quantifier. We cannot assume anything about it other than the condition in the statement to be proven, the hypothesis: $d((x,x_0), (y, y_0)) < \delta$, or for our taxicab metric:
$$|x - x_0| + |y - y_0| < \delta = \delta_f + \delta_g$$
From this you need to show that $|f(x) - f(x_0)| + |g(y) - g(y_0)| < \epsilon$
It would be easy if we knew that $|f(x) - f(x_0)| < \frac \epsilon 2$ and $|g(y) - g(y_0)| < \frac \epsilon 2$. But we don't know that yet. What we know is that these hold when $|x - x_0| < \delta_f$ and $|y - y_0| < \delta_g$. But we don't know that, and in fact by your choice of $\delta$, they don't even have to be true.
We know $$|x - x_0| + |y - y_0| < \delta_f + \delta_g$$ But any shortfall of $|x - x_0|$ below $\delta_f$ allows $|y - y_0|$ to be greater than $\delta_g$ and since $x,y$ are arbitrary, $y$ for which this is true will be included.
So $|x - x_0| + |y - y_0| < \delta_f + \delta_g$ is not sufficient to guarantee that $|x - x_0| < \delta_f$ and $|y-y_0|<\delta_g$, and the proof is stuck. Now consider what would have happened if you had chosen $\delta = \min\{\delta_f, \delta_g\}$ instead.
