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.

Take just one exponential. Do you see that the set $e^x\le c$ is unbounded for all $c>0$? Use the same idea for several variables.

Proof: the exponential function is increasing, i.e. for any $x'\le x$ we have $e^{x'}\le e^x$. Hence,
$$
e^{x_1'}+e^{x_2'}\le e^{x_1}+e^{x_2}\le c
$$
is satisfied whenever $x_1'\le x_1$ and $x_2'\le x_2$ for $(x_1,x_2)$ in the set. It makes it possible for $x_1'$ and $x_2'$ to go to $-\infty$ within the set, hence, it is unbounded.
