Show that $\lim{\sup{A}}$ is finite iff $A$ has a limit point Definition. Let $A$ be a nonempty subset of $\mathbb{R}$. $x \in \mathbb{R}$ is called an almost upper bound of $A$ if there are only finitely many $y \in A$ for which $y \geq x$. Similarly we define almost lower bounds. Define $\lim \sup{A}$ to be the infimum of all almost upper bounds of $A$ and $\lim \inf{A}$ to be the supremum of all almost lower bounds of $A$.

Show that $\lim{\sup{A}}$ is finite iff $A$ has a limit point. Similarly show that $\lim{\inf{A}}$ is finite iff $A$ has a limit point.

I don't see how this statement makes sense. If $A = \{1,2,3\}$ isn't $\lim{\sup{A}}$ finite but doesn't have a limit point?
 A: If the set $A$ is finite, then every real number is an almost upper bound. The reason is that for any $x$, it is true that only finitely many members of $A$ exceed or are equal to $x$, since $A$ is finite! So $\limsup A$ is the inf of the entire set of reals, which is not finite. Similarly $\liminf A$ is not finite. So your example is not a counterexample.
However, the statement you are trying to prove appears to be true under the additional assumption that $A$ is bounded.
Proof: Spse $a:=\limsup A$ is finite. Let $\epsilon>0$. Since $a$ is the inf of the set of AUBs, the number $a-\epsilon$ is not an AUB. So there are infinitely many members $y$ of $A$ for which $y\ge a-\epsilon$. Moreover, since only finitely many members of $A$ exceed $a+\epsilon$, there exists a member $y$ of $A$ such that $a-\epsilon<y<a+\epsilon$. Since $\epsilon$ was arbitrary, we've shown that $a$ is a limit point of $A$. (This direction doesn't require $A$ to be bounded.)
To prove the converse, suppose $A$ is bounded. Let $a$ be the sup of the set of limit points of $A$, so that $a$ is finite. There are infinitely many members of $A$ that exceed $a-1$ but only finitely many that exceed $a+1$. It follows that the set of AUBs is nonempty and bounded below. Therefore the inf of the set of AUBs is finite. 
A: I think that the statement is false. Consider the set
$$
A=\{1/n\,|\, n\in\mathbb N\setminus\{0\}\}\cup\mathbb N
$$
It has a limit point (i.e. 0) but the $ \limsup$ is not finite.
A: The statement is false . Consider the interval $(0,\infty)$. The set of almost upper bounds of $(0,\infty)$ is empty. And we know  inf $  \varnothing =\infty$
A: I like your definition of limit superior of a set A. I have read the concept of limit superior in the context of limit of functions and limit of sequences. But it appears that your definition takes a general view and defines it for a set of numbers. The usual definition matches your definition if we apply your definition (with slight modifications) on the set of values taken by the function or sequence.
However the problem you mention is not correct. Your problem statement is true only when $A$ is bounded above (if we are dealing with $\limsup A$). Similar statement holds for $\liminf A$ if $A$ is bounded below. I use this assumption in the proof which follows.

Let $A$ be a non-empty set of real numbers which is bounded above. We first prove that if $A$ has a limit point $c$ then $\limsup A$ is finite. Now here is the catch! When we say that $A$ has a limit point $c$ then we don't necessarily mean that $c \in A$, but rather that there is some number $c$ which is a limit point of $A$. It is easy to see that the set $B = \{x \mid x \text{ is a limit point of }A\}$ is non-empty and has a greatest member $M$. We show that $\limsup A = M$. This is done via proving the following statements:


*

*If $\xi > M$ then $\xi$ is an almost upper bound for $A$.

*If $\xi < M$ then $\xi$ is not an almost upper bound for $A$.


Clearly if $\xi > M$ then $\xi$ is not the limit point of $A$ because $M$ by definition is the greatest limit point of $A$. It is now clear via Bolzano-Weierstrass Theorem that there can be only finitely many members of $A$ which are greater than or equal to $\xi$. Thus $\xi$ is an almost upper bound of $A$.
If $\xi < M$ then we can see that there are infinitely many points of $A$ which are so near to $M$ as to exceed $\xi$ and hence $\xi$ is not an almost upper bound of $A$.
Thus we have proved that if there is a limit point $c$ of $A$ then $\limsup A$ exists as a finite real number.
Let's prove the converse. Suppose $\limsup A = M$. Then we just need to prove that $M$ is also a limit point of $A$. This is almost obvious. Let $\epsilon > 0$ be given arbitrarily. Then there is an almost upper bound $M'$ of $A$ with $M \leq M' < M + \epsilon$ so that only finitely many members of $A$ are greater than or equal to $M'$. Further $M - \epsilon < M$ and hence $M - \epsilon$ is not an almost upper bound of $A$. It follows that there are infinitely many members of $A$ which are greater than or equal to $M - \epsilon$. It thus follows that there are infinitely many members of $A$ which lie in $[M - \epsilon, M + \epsilon]$ and since $\epsilon > 0$ is arbitrary it follows that $M$ is a limit point of $A$.
