functions between metric spaces: How much freedom do I have, when I choose the metrics to proof continuity of a function? $\epsilon-\delta$-criteria Let $X, Y$ be metric spaces with the metrics $d_X: X\times X \rightarrow \mathbb{R}$ and $d_Y: Y\times Y \rightarrow \mathbb{R}$.
A funtion $f:X\rightarrow  Y$ is continious at the point $a$, if for any $\epsilon >0$, there exists an $\delta >0$ such that for every $x\in X$:
$d_X(x,a)<\delta \Rightarrow d_Y(f(x),f(a))<\epsilon $
Assuming that the metrics $d_X$ and $d_X'$ are equivalent..
The same assumption for $d_Y$ and $d_Y'$.
Let $(X,d_X),(X,d_X'),(Y,d_Y)$ and $(Y,d_Y')$ be metric spaces.
If I showed that $f$ is continuous at a point $a\in X$ with respect to $(X,d_X)$ and $(Y,d_Y)$, I think that I can conclude that $f$ is also continuous in the metric spaces:

*

*$(X,d_X)$ and $(Y,d_Y')$


*$(X,d_X')$ and $(Y,d_Y)$


*$(X,d_X')$ and $(Y,d_Y')$
Reasoning: A funtion $f:X\rightarrow  Y$ is continious at the point $a$, if for any $\epsilon >0$, there exists an $\delta >0$ such that:
$x\in U_{\delta,d_X}(a) \Rightarrow f(x)\in U_{\epsilon ,d_Y}(f(a))$
Case 2.)
The metrics $d_X$ and $d_X'$ are equivalent, so $U_{\delta,d_X}(a)$ is also open in $(X,d_X')$, so that one can find a $\delta'>0$ such that $U_{\delta',d_X'}(a)\subset U_{\delta,d_X}(a)$.
$x\in U_{\delta',d_X'}(a)\Rightarrow x\in U_{\delta,d_X}(a)\Rightarrow f(x)\in U_{\epsilon ,d_Y}(f(a))$
Case 1.)
Let $\epsilon >0$ be given and the implication $x\in U_{\delta,d_X}(a) \Rightarrow f(x)\in U_{\epsilon ,d_Y}(f(a))$ be true for a $\delta>0$.
The metrics $d_Y$ and $d_Y'$ are equivalent.
Here it is harder for me to argue for this implication:
$x\in U_{\delta,d_X}(a)\Rightarrow f(x)\in U_{\epsilon ,d_Y}(f(a))\Rightarrow f(x)\in U_{\epsilon ,d_Y'}(f(a))$
Because $\epsilon$ is fixed.
My main question is best explained with an example:
$$f:\mathbb{R}^2\rightarrow \mathbb{R}, \\
f(x,y)=\begin{cases}0 & (x,y)=(0,0) \\ \frac{2x^2y}{x^2+y^2} & (x,y)\neq (0,0) \end{cases}$$
My proof that this function is continuous in $(x,y)=(0,0)$:
I use the Euclidean metric $d_e(x,y)=\sqrt{x^2+y^2}$ and $d(f(x,y),f(x',y'))=|f(x,y)-f(x',y')|$.
Let $\epsilon >0$ be given.
Choose $\delta =\epsilon$, for any $(x,y)\neq 0$ with $d_e(x,y)=\sqrt{x^2+y^2}<\delta$ follows:
$$ |f(x,y)-f(0,0)|=|f(x,y)|=|\frac{2x^2y}{x^2+y^2}|=2|xy||\frac{x}{x^2+y^2}|\leq |x^2+y^2||\frac{x}{x^2+y^2}|=|x|\leq |x|\sqrt{1+\frac{y^2}{x^2}}= \sqrt{x^2+y^2}<\delta =\epsilon$$
I know that every norm $q$ induces a metric and every norm in $\mathbb{R^n}$ with $n\in \mathbb{N}$ is equivalent.
Since $d_e$ is induced by the Euclidean norm, can I now conclude that $f$ is continuous at the point $(0,0)$ in every metric space with an metric $d_N: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}, (x,y)\mapsto q(x,y) $ that is induced by a norm $q$ ???
So if I need to proof that $f$ is continuous with respect to the metric spaces $(X,d_X)$ and $(Y,d_Y)$, I could just show that $f$ is continuous with respect to the metric spaces $(X,d_X')$ and $(Y,d_Y')$, if $d_Y',d_Y$ and $d_X,d_X'$ are equivalent?
As an example I could have used the Manhattan norm to proof that $f$ is continuous with respect to $(\mathbb{R}^2,d_e)$ and $(\mathbb{R}^2,d)$:
Choose $\delta =\epsilon$, for any $(x,y)\neq 0$ with $d_1(x,y)=|x|+|y|<\delta$ follows:
$$ |f(x,y)-f(0,0)|=|f(x,y)|=|\frac{2x^2y}{x^2+y^2}|=2|xy||\frac{x}{x^2+y^2}|\leq |x^2+y^2||\frac{x}{x^2+y^2}|=|x|\leq |x|+|y|<\delta =\epsilon$$
 A: For metric spaces $(X,d)$ and $(Y,\rho)$, and for a function $f:X\to Y$ and a point $x_0\in X$, $f$ is continuous at $x_0$ if and only if whenever $(x_n)_{n=1}^\infty\subset X$ satisfies $\lim_n x_n=x_0$, then $\lim_n f(x_n)=f(x_0)$. This holds for metric spaces, but not all topological spaces.
Equivalent metrics have the same convergent sequences, so it's easy to see using the sequential criterion from the previous paragraph that continuity of $f:(X,d)\to (Y,\rho)$ at $x_0\in X$ is equivalent to continuity of $f:(X,d')\to (Y,\rho')$ at $x_0$, where $d,d'$ are equivalent on $X$ and $\rho, \rho'$ are equivalent on $Y$. If the preceding paragraph holds for sequences which are $d'$-convergent in $X$ and $\rho'$-convergent in $Y$, then it also holds for sequences which are $d$-convergent in $X$ and $\rho$-convergent in $Y$, since $d$-convergence is the same as $d'$-convergence in $X$ and $\rho$-convergence is the same as $\rho'$-convergence in $Y$.
If you want to prove the equivalence using only the $\delta$-$\epsilon$ condition without using the sequential characterization, you'll want to start with an $\epsilon>0$. By equivalence of $\rho$ and $\rho'$, there exists $\epsilon'>0$ such that $U_{\epsilon',\rho'}(f(x_0))\subset U_{\epsilon,\rho}(f(x_0))$.  By $d'$-$\rho'$ continuity, there exists $\delta'>0$ such that $f(U_{\delta',d'}(x_0))\subset U_{\epsilon',\rho'}(f(x_0))$. By equivalence of $d$ and $d'$, there exists $\delta>0$ such that $U_{\delta,d}(x_0)\subset U_{\delta',d'}(x_0)$.
A: You are already done!
There is some nested ball of radius r:
$$U_{r ,d_{Y}}(f(a)) \subset U_{\epsilon ,d'_{Y}}(f(a)) $$
So by continuity of $f$ at $a$ take the appropriate $\delta$.
