How to prove $p:\mathbb{R^{2}}\to \mathbb{R}$ where $p((x,y))=x$ is continuous How can we prove that the projection map $p\colon\mathbb{R}^2\to\mathbb{R}$, $p(x,y)=x$, is continuous?
This is a very simple question, but I have only had to prove continuity from $\mathbb{R} \to \mathbb{R}$ before, so I would greatly appreciate it if someone could help me how to do this correctly, can we do it using the open balls
Where $c=(x_1,x_2)\in \mathbb{R^{2}}$
$B_{\delta}(c)=\{(x,y): \| (x,y)-c \|<\delta\}$.
$B_{\epsilon}(p(c))=\{x:|x-p(c)|<\epsilon\}$.
And show $\forall c \in \mathbb{R^{2}}$ if
 $a\in B_{\delta}(c) \implies p(a)\in B_{\epsilon}(p(c)) $.
 A: Let $c=(x_1,x_2)$, and let $\epsilon\gt 0$. You want to show that there exists $\delta\gt 0$ such that if $\lVert (x,y)-(x_1,x_2)\rVert \lt \delta$, then $|x-x_1|\lt\epsilon$.
Let $\delta=\epsilon$. Since
$$\lVert (x,y)-(x_1,y_1)\rVert = \sqrt{(x-x_1)^2 + (y-y_1)^2} \geq \sqrt{(x-x_1)^2} = |x-x_1|,$$
then if $\lVert(x,y)-(x_1,y_1)\rVert\lt\epsilon$, then $|x-x_1|\lt\epsilon$. Therefore, $p$ is continuous. 
A: I was going to give Arturo's answer, but I'll just mention that if we endow $\mathbb{R}^2=\mathbb{R}\times\mathbb{R}$ with the product topology, then the projection map $p:\mathbb{R}^2\to\mathbb{R}$ onto the first coordinate is continuous essentially by definition.
One can prove that the product topology on $\mathbb{R}^2$ is the same as the metric topology induced by the metric $$d((x_1,y_1),(x_2,y_2))=\|(x_2-x_1,y_2-y_1)\|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},$$ so that whatever answer we get by thinking about $\delta$'s and $\epsilon$'s will be the same as the answer we get by thinking about open sets.
A: To add on Zev and Arturo's answers, one can just verify that the preimage of an interval is an open subset of the plane.
In particular, $p^{-1}[(a,b)]=\{\langle x,y\rangle\mid a<x<b, y\in\mathbb R\}=(a,b)\times\mathbb R$, which is an open set in the product.
Therefore the preimage of a basic open set in $\mathbb R$ is an open set in $\mathbb R\times\mathbb R$.
