Is $ A = \left\{ \frac{1}{n} \mid n \in \mathbb{N} \right\} $ as a subspace of $(0, +\infty)$ closed? Is $ A =  \left\{ \frac{1}{n} \,\middle|\, n \in \mathbb{N} \right\} $ as a subspace of $( 0, +\infty) $ closed? Also is it compact? Can I use the definition that a set is compact if it is closed and bounded subset of $\mathbb{R}$ in this case because I'm looking at a subspace of $\mathbb{R}$?
 I would say that it is closed because its complement in $(0, +\infty)$ is open. Then it would alse be compact because it is bounded.
 A: As you have correctly observed, $A$ is closed (as a subspace of $(0,\infty)$) since its complement is an open subset of $(0,\infty)$.
On the other hand, $A$ is not compact.  In particular, the sequence $(\frac 1n)_{n \in \Bbb N}$ is a sequence in $A$ that has no convergent subsequence (since $0$, the sequence's limit in $\Bbb R$, is not in $A$).
A: One of the other answers has already used the sequential definition of compactness, so here is a way to think about why $A$ not compact using the open cover definition. For each $\frac{1}{n}$, we can consider the open ball of radius $\epsilon = \min \big\{ \frac{1}{n} - \frac{1}{n+1}, \frac{1}{n-1} - \frac{1}{n}\big\}$. The collection of these open balls forms an open cover of $A = \big\{ \frac{1}{n} : n \in \mathbb{N}\big\}$, yet each open ball contains only one element of $A$. Therefore, this (infinite) open cover does not admit a finite subcover, so $A$ is not compact.
A: It's closed as its complement is just
$$\bigcup_{n\in \mathbb{N}}\left(\frac1{n+1},\frac1n\right)\cup(1,+\infty)$$
which is a union of open sets.
For compactness, see @Omnomnomnom's answer.
A: Unfortunately, when we consider $(0,\infty)$ as a topological space in its own right, as a subset of $\mathbb R$ under the subspace topology, it is not true that if a set is bounded and closed, then it is compact.
You may want to go back to the proof of the Heine-Borel theorem (which states that the above is true for $\mathbb R$), and see where it fails for the above case.
Therefore, compactness may be checked/negated either via definition (open cover) or using the fact that in a metric space, this is equivalent to the notion of "sequential compactness". Let us put these definitions side by side :
Usual : Every open cover has a finite subcover.
Sequential compactness : Every sequence has a convergent subsequence, with limit inside the set.
To check if a set in a metric space is compact, any one of the above criteria may be used.

With this in mind, sequential compactness is readily seen to not hold for the given set, since $\{\frac 1n\}$ is a sequence in $A$, which has no convergent subsequence in $A$, since the sequence, and therefore every subsequence, converges to $0$ in the metric space topology , but $0 \notin (0,\infty)$, so the fact is that no subsequence of $A$ is convergent. Therefore, $A$ is not compact.
For using the other definition, we note that open sets in $(0,\infty)$ are the intersection of $(0,\infty)$ with usual open sets in $\mathbb R$. So, since $A$ is not compact in $\mathbb R$ (it does not contain the limit point $0$, so is not closed), so there is an open(in $\mathbb R$) cover of $A$ having no finite subcover. Taking the intersection of all  these open sets with $(0,\infty)$ gives us an open (in $(0,\infty)$) cover of $A$ which has no finite subcover. Thus, even via a direct means of attack we can contradict compactness of $A$.

Note that $A$ is closed in $(0,\infty)$, since it has no limit points, and $A$ is bounded. However, $A$ is not compact. 
