To determine whether range of f is closed , connected etc Let $E= \{ (x,y) : |x| + |y| \leq 1 \}$  . Define $f :  E \to \mathbb R$ by 
$f(x, y) = x + y / 1 + x^{2} + y^{2} $
Then range of $f$ is 
A . Connected open set
B . Connected closed set 
C.  Bounded open set
D . Closed and unbounded set
I tried putting values of $x$ and $y$ in $f$, like $(0,0)$  or $(1,1)$. By doing this for many pairs I concluded that the range lies in $(-1,1)$. But I do not know a concrete method for such questions. If I am right then the set is not closed as $1$ is not in the set, but how do I check for other options. Help will be appreciated. Thanks.
 A: Since $f$ is a continuous function and $E$ is a compact set being a closed and bounded subset of $\mathbb{R}^2$, $f(E)$ is compact. Hence it is closed. Similarly; since $E$ is connected, $f(E)$ is connected. 
For the proof that a continuous function maps compact sets to compact sets, connected sets to connected sets: (on page 2, 3.1 and 3.2)
http://www.math.jhu.edu/~fspinu/405/405-continuity%20thms.pdf
A: The image (range) of $f$ is
$$ f(E) = \{ f(x,y) : |x|+|y|\leqslant 1\} =\left\{ \frac{x+y}{1+x^2+y^2} : |x|+|y|\leqslant 1\right\}. $$
First, you should get an idea of what the set $E$ looks like - if you draw a picture, you'll see that it is the closed unit ball in the $L^1$ norm (sometimes called the "taxicab" distance). So, $E$ is a closed and bounded set of $\mathbb R^2$, hence it is compact. Further, $E$ is connected (in fact it is convex and path-connected, but that is not necessary here).
It's easy to see that $f$ is continuous, as it is a rational function and the denominator is always $\geqslant 1$ (as $x^2,y^2\geqslant 1$). Therefore $f(E)$ is both compact and connected. Being compact, $f(E)$ is closed and bounded. So from your choices, B ("connected closed set") would be correct. Remember though that $E\subset\mathbb R^2$ while $f(E)\subset\mathbb R$.
