# Why are both of the conditions of Bolzano -Weierstrass theorem necessary?

One version of the Bolzano-Weierstrass theorem states that:"An infinite bounded set of real numbers has at least one limit point." From the definition of limit point, it's easy to see that a finite set can not have a limit point. However, the requirement of boundedness is not so obvious. Before giving the proof of the theorem, my textbook uses the set of integers as an example, says that "the set of integers, though infinite, has no limit point because it's not bounded." How to understand this sentence?

• Do you know what a limit point is? – ajotatxe May 8 '15 at 14:13
• The definition of limit point given in my textbook is as follows: "if S is an infinite set of real numbers, a real number x is called a limit point of S if every ϵ-neighborhood of x contains infinitely many elements of the set S." – Song Wang May 8 '15 at 14:17
• The textbook is misleading. The set of integers has no limit point, but not simply because it isn't bounded. There are unbounded sets that have limit points. If only textbook authors knew how much difficulty an awkward sentence can cause... – Reinstate Monica May 8 '15 at 14:40
• @ Solomonoff's Secret, please show me an example. – Song Wang May 8 '15 at 14:43
• @SongWang Every real number is a limit point of $\mathbb{R}$. Another example is $\mathbb{Z} \cup \{3, 3.1, 3.14, 3.141, \ldots\}$, which has limit point $\pi$. – Reinstate Monica May 8 '15 at 14:46

Clearly, in order to have any kind of limit point, you need an infinite number of elements. But just because you have an infinite number of elements, doesn't mean you'll have a limit point, which is what the example you mentioned shows: $\mathbb Z$ is infinite and yet has no limit point.
Any convergent sequence (i.e., one that converges to a finite value) must be bounded. Proof: Let $L$ be the limit of the sequence. By definition all the elements of the sequence are within distance $1$ from $L$ as long as one goes far enough in the sequence. Thus the entire sequence must be bounded by the largest (in absolute value) number determined by some initial segment of the sequence, and $|L\pm 1|$.