How to show that $\{(x_1,x_2) \in \mathbb{R}^2 | \exp(x_1) + \exp(x_2) \leq c\}$ is unbounded if $c > 0$ Define $f(x) := \exp(x_1) + \exp(x_2) $
Let the sublevel set be given by $\{x = (x_1,x_2) \in \mathbb{R}^2 | \exp(x_1) + \exp(x_2) \leq c\}$
A plot of this function along with its contour is given as:

Clearly, the sublevel sets are unbounded for every $c>0$. But how do you prove this? 
I am thinking let $x_2 = x_1 + a, a \in \mathbb{R}$, then we have $ \exp(x_1) + \exp(x_2) \leq c \implies \exp(x_1)(1+\exp(a)) \leq c \implies \exp(x_1) \leq k = \frac{c}{1+\exp(a)}$ and the latter inequality is satisfied by an uncountably many $x_1$, this way we cannot place a ball large enough to contain this set. But then this is only on one line (<- never mind). Very rough arguments here.

Since  $\lim_{x\to -\infty} e^x =0$ it follows that for any $\frac{c}{2} \in \mathbb {R}^+$ there is some $a$ such that $e^x\leq \frac{c}{2}$ for all $x<a$ so $e^x + e^y \leq c$ for $x<a$ and $y<a$
This implies that the sublevel set $f^{-1}((-\infty,c])$ contains the set $\{(x,y) \in \mathbb {R}^2|x<a , y<a\}$ which is unbounded so $f^{-1}((-\infty,c])$ is unbounded.  
