Is the set $H= \{x^2: x \in \Bbb R\}$ an open or closed set? 
Is the set $H= \{x^2: x \in \Bbb R\}$  open or closed in $\Bbb R$ ?

I believe this set is open since every arbitrary neighborhood contains an element of our set. $x$ is an element of the reals. I'm not sure if it is closed. How can we prove it either way?
 A: Your intuition about open set is wrong! For definition, see this wiki link.
Here $$H=\{x^2: x \in \Bbb R\}=[0, \infty)$$ so it is closed, since it's complement $(-\infty, 0)$ is open. H is not open, since any nbd of zero does not entirely contained in H
The other way to see H is not open as follows:
Since it is already closed, so if in addition it were open, then the real line becomes a disconnected set, which is false.
A: It should be clear that $H$ is the set of nonnegative reals. In other words, $H=[0,\infty)$. For any $\varepsilon>0$, a neighborhood of radius $\varepsilon$ around $x=0$ is partially outside $H$. This means that $H$ is not open since for a set to be open, by definition, all points in the set must live within a neighborhood which is entirely contained in the set.
You can use this same argument to show that $H^c$ (the complement of $H$ in $\mathbb{R}$) is open, hence $H$ is closed. Equivalently, what are the limit points of $H$? A set is closed if it contains all of its limit points, and exploring this method would be a good exercise.
