Is $A = f\mathbb{R^2}$ complete? $f:\mathbb{R^2}\to\mathbb{R^3}$, $f(x,y) = (x,y,x^2-y)$ My question is pretty much what it says in the headline.
Is $A = f\mathbb{R^2}$ complete, where $f:\mathbb{R^2}\to\mathbb{R^3}$, $f(x,y) = (x,y,x^2-y)$, $(x,y)\in\mathbb{R^2}$.
$f\mathbb{R^2}=$$\{ f(x,y), (x,y) \in \mathbb{R^2} \}$
My initial thought is that as $f\mathbb{R^2}$ is $f$:s image (i.e. "set of values"), $f\mathbb{R^2}=\mathbb{R^3}$, but I'm not sure, and I'm not sure how could I show this. It is not intuitively clear to me because $x^2-y$, i.e. the third component of $f$ depends on the other two components. 
Anyway if it was so that $f\mathbb{R^2} = \mathbb{R^3}$, we know that $\mathbb{R^3}$ is complete, so the image in question would be complete as well. 
Any comments on how I tried to approach this. Any suggestions on other approaches? 
Thanks in advance. 
 A: Thanks for your help, I'm pretty sure I got it now. And I think I understood the idea of homeomorphisms better. 
Yes, thank you for pointing out the two different concepts. I still think it is image I was trying to refer to. I think my course book uses unorthodox notations every now and then.
According to my course book
Let $f:X \to Y$ and $A$ subset of $X$. 
Now $fA = \{ f(x),x \in A \}$. This is what the course book calls the image of f.
So now to answer the question whether $fA$ is complete, I can define the functions
$f_1: X \to fX$, $f(x,y) = (x,y,x^2-y)$ and $g: fX \to X$, $g:(x,y,z)=(x,y)$. [Note that here $z \in fX$, so that it obeys $z=x^2-y$]
So $f_1$ and $g$ are continuous. 
Moreover $g (f(x,y)) = (x,y)$ and $f(g(x,y,z))=(x,y,x^2-y)$, the identity functions. (At least, so they are called in my book). 
So now we see that $f_1$ is a homeomorphism between $X$ and $fX$, where $X=\mathbb{R^2}$. 
And, as $\mathbb{R^2}$ is complete, so must $f\mathbb{R^2}$ be also complete.  
A: Let $X$ and $Y$ be metric spaces.  If $V \subset X$, define the graph of a function $f: V \to Y$ to be the set $\big\{(x, f(x) \; : \; x \in V \big\} \subset X \times Y$.  $X \times Y$ is itself a metric space under the metric $d((x_1,y_1),(x_2, y_2)) = d(x_1, y_1) + d(x_2, y_2)$, or $\sqrt{d(x_1, y_1)^2 + d(x_2, y_2)^2}$ if you prefer.  (Any of the $p$-norms are topologically equivalent, i.e. the same sets are closed and open in the product metric space.)
Proposition. Let $V \subset X$ be closed, and let $f : V \to Y$ be continuous.  Then the graph of $f$ is closed in $X \times Y$.
Proof.
Let $(x,y)$ be a limit point of the graph of $f$.
Then there exist $x_1, x_2, \ldots$ such that $(x_n, f(x_n)) \to (x,y)$.
In particular, this means $x_n \to x$ and $f(x_n) \to y$.


*

*Since $V$ is closed, $x_n \to x$ implies $x \in V$.

*Since $f$ is continuous, $x_n \to x$ implies $f(x_n) \to f(x)$.  Since $f(x_n) \to y$ also, we must have $f(x) = y$.


Therefore $(x,y)$ is the graph of $f$, and so the graph of $f$ is closed. $\square$

In your case, we have that $\mathbb{R}^2$ is closed, so the graph of the function $g(x,y) = x^2 - y$ is a closed subset of $\mathbb{R}^3$.  Since $\mathbb{R}^3$ is complete, this closed subset must additionally be complete.
