Determining if the set $\{e^{-x} : x\geq 0\}$ is compact or not.

Question

Decide if the set $\{e^{-x} : x\geq 0\}$ is compact or not. If not compact give a counter example.

Reasoning:

So it is known that if a set is compact it is closed and bounded. The two possible trouble points of this set are the endpoints. If $x = 0$ then $e^{-x} = 1$ this would serve as an upper bound and would also highlight that this limit point is in the set. I'm having trouble with reconciling the other possible point $0$.

So $0$ is definitely a lower bound for this set, but I'm not sure whether I can conclude that it is in the set. The reason for this is that a set is considered closed if it contains all of its limit points. There is definitely a sequence converging to $0$, and I "feel" that $0$ "should" be considered in the set, but I'm having an issue with dealing with the limit operator on the sequence: $$\lim_{x \rightarrow \infty} e^{-x} = 0$$

Does the fact that I am taking the limit still allow me to arrive at the conclusion that $0 \in \{e^{-x} : x\geq 0\}$?

Answer 1

$0$ is a limit point of the set but exponential is always positive so set is not closed.

Answer 2

$0\in \{e^{-x}:x\geq 0\} \iff \exists x\geq0:e^{-x}=0$. Since $\forall x: e^{-x}> 0$, we can conclude that 0 is not in the set.

Answer 3

If you think of the graph of $e^{-x}$, you'll see that $\{e^{-x} \mid x \geq 0\} = (0,1]$, which is not compact, as it is not closed.