Jtdylktuy

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.

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.

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 $A$).

It's closed as its complement is just
$$bigcup_{nin mathbb{N}}left(frac1{n+1},frac1nright)cup(1,+infty)$$
which is a union of open sets.

For compactness, see @Omnomnomnom's answer.

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.