Proving a mapping from a higher dimension to a lower dimension is continuous? How does one show a mapping is continuous from a higher dimension to a lower dimension. For example:
Let f : $\mathbb{R}^2$ $\rightarrow$ $\mathbb{R}$, (x,y) $\rightarrow$ x. How would you show this is continuous.
My attempt was to use the definition of continuous:
That is: f is continuous at $x_0 \in$ its domain if and only if for every $\epsilon$ > 0, there exists a $\delta$ > 0 such that for all x $\in$ its domain A, d(x,$x_0$) < $\delta$ implies $\gamma$(f(x),f($x_0$)) < 0.
So set y = $y_0$ and then the proof should be vascuous. However, I feel that there is something wrong with this approach.

Let $z_0=(x_0,y_0)$ be an arbitrary point in the domain. Let $\epsilon>0$.
We need to exhibit $\delta>0$ such that for all $z=(x,y)$ with $d(z,z_0)<\delta$ we have $d(f(z),f(z_0))<\epsilon$. Here we have the Euclidean metric on $\Bbb R^2$, i.e., $d(z,z_0)=\sqrt{(x-x_0)^2+(y-y_0)^2}$ and the standard metric on $\Bbb R$, i.e., $d(f(z),f(z_0))=|x-x_0|$.
As $$\sqrt{(x-x_0)^2+(y-y_0)^2}\ge \sqrt{(x-x_0)^2}=|x-x_0|$$
we see that we can simply pick $\delta=\epsilon$.
