Real Analysis: Compact Sets I'm working on a general real analysis problem involving compact sets. I was given these two sets: 
$$A = \left\{0, 1, \frac{1}{2}, \frac{1}{3}, \dots , \frac{1}{n}, \dots\right\}\text{ and }
B = \left\{1, \frac{1}{2}, \frac{1}{3}, \dots , \frac{1}{n}, \dots\right\}$$
I'm supposed to figure out which set is compact and which is not, and explain why. My intuition tells me that $A$ is compact and $B$ is not compact. But I'm not sure why. Could someone give me a proof of (or show me) why $A$ is compact and $B$ is not? 
Help is greatly appreciated. 
 A: 1) $A = \{0 \} \cup \{1/n \}$ is compact since if we take any open cover of $A \subset \bigcup_\alpha U_\alpha$ there exists $\alpha_0$ such that $0 \in U_{\alpha_0}$, but then there are only finitely many points of $A$ outside this neighborhood to cover.  Hence there exists a finite sub cover so $A$ is a compact set.
2) $B$ is not compact because it is a subset of the real line that is not closed.
A: Hint: For $A$, show that any open cover admits a finite subcover. For $B$, notice that the sequence $(1/n)$ has no limit point in $B$.
A: Compact sets are closed.  A limit of a certain sequence in $B$ does not converge to any point in $B$, so $B$ is not closed.
A: Notice that $A$ has all its limit points, so it is closed and it's also bounded. However $B$ does not satisfy that definition.
A: in any finite dimensional space every closed and bounded set is compact. and A is closed since it has all it's limit point (only one limit point '0') and also it is bounded. So it is a compact set but B is not.
A: $A$ is closed and bounded but $B$ is not closed (0 is not in $B$, which is a limit point of $B$).
Hence $A$ compact but $B$ does not.
Start by proving only 0 is the limit point of $A$.
A: Def of compact set is closed and bounded. Here A and B are bounded.we show that closed  there limit    is exist and limit point is 0.but A is belongs to 0 and B doesn't  belongs to 0 . So A is closed but B is not. hance A is compact but B is not.
