I want to show that the set $\{e^{-x}: x > 0, x\in \mathbb{R}\}$ is neither open or closed rigourously. I have the idea intuitively which I will describe for each scenario below, but before that two definitions:
Defn 1): A set $A$ is said to be open if there exists an $\epsilon$ -neighbourhood around every point $x \in A$ s.t $B_{\epsilon}(x) \subset A$
Defn 2): A set $A$ is said to be closed if the set $A$ contains all of its limit points.
Proof Idea
i) not open:
If you draw the graph of $f(x) = e^{-x}$ you get the inverse exponential function. Now picking any point on this graph you can "see" that any $\epsilon$– neighbourhood around a point will contain points not in the set, in particular any point to the left or right of the point $x \in \{e^{-x}: x > 0, x\in \mathbb{R}\}$
ii) not closed:
This one I am a little bit more shakey on. The point $0$ is in the set if you take the limit as $x \rightarrow \infty$, but if you take $x \rightarrow 0$ you head towards $+\infty$. I'm not sure how to classify this. Since this set extends over all of $\mathbb{R}$ I want to include it, but from past readings I think it might not be the case.
My Problems: These are my ideas, so how would I make them rigourous? I can't think of anything more beyond this sort of paragraph form of explanation. Possibly I'm too caught up in wanting to use symbolic notation…
Best Answer
For not open, your proof does not work because you just have the points in $\Bbb R$, not the points $(x,f(x))$ in $\Bbb R^2$. Can you represent the set in a much simpler way? What is the range of the function? In fact the set is open.
For not closed, you want to find a limit point not in the set. $1$ works because of the restriction $x \gt 0$. Find a sequence in the set that converges to $1$.