how to show that f is continuous?

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I'm trying to learn how to do delta-epsilon proofs of continuity from "Introduction to Topology" by Bert Mendelson. The problem I'm working on is this:

Define the function $f:R^2\rightarrow R$ by $f(x_1,x_2)=x_1+x_2$. Prove that $f$ is continuous, where the distance function on $R^2$ is either $d$ or $d'$.

And then d and d' have been previously defined in the text. (Right now I'm working on $d'$, which is the euclidean distance function.)

My question is, does my $\delta$ have to be constant, or can it be a function of $(x_1,x_2)$?

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Your $\delta$ can be a function of $(x_1,x_2)$, since you are trying to prove coontinuity on each point of the function's domain.

The case where $\delta$ is constant (same for all points) is called uniform continuity and is much stronger than simple continuity.

Uniform Continuity

Edit: To get an intuition for both cases take as example functions from $\mathbb{R}$ to $\mathbb{R}$ like $f(x)=x$ and $g(x)=\exp{x}$, both are continuous.

Now given $\epsilon>0$ you can take $\delta=\frac{\epsilon}{2}$ and for any point you have $|f(x\pm\delta)-f(x)|<\epsilon$ showing the uniformity of $f$.

While for $g$ you can't find such $\delta>0$ for a given $\epsilon>0$ since as you increase $x$ the difference $|g(x\pm\delta)-g(x)|=|g(x)||g(\pm\delta)-1|$ increases unbounded.

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The definition of continuity you're using is the following: A function $f : \Bbb{R}^2 \to \Bbb{R}$ is continuous at the point $P_0 \in \Bbb{R}^2$ when given any $\epsilon >0$, there exists a $\delta > 0$ such that if $P \in \Bbb{R}^2$ with $d'(P,P_0) < \delta$, then $|f(P)-f(P_0)|<\epsilon$. When reading this definition, notice that before $\delta$ is introduced, you already have a function $f$, a point $P_0$, and some "tolerance" $\epsilon >0$, and since you only need to provide the existence of a particular $\delta > 0$, your choice is allowed to depend on $f$, $P_0$, and $\epsilon$. In particular, if $P_0 = (x_1,x_2)$, then $\delta$ can certainly depend on $x_1$ and $x_2$.