# Any finite set is compact; what exactly is a finite set?

Any compact set is finite. Assume the sets are in $\mathbb{R}$ Since $A = [0, 1]$ is compact, it is also finite. As for $B = (0, 1)$, it is not compact, so it is infinite. However, how is it infinite? There are infinitely many points between 0 and 1? But aren't there infinitely many points between 0 and 1 in set $A$ as well?

Also, for the set $C = \left \{ 1 + \frac{1}{n}, n \in \mathbb{N} \right \}\cup \left \{ 0 \right \}$. It is bounded and closed, thus compact. However, how is it finite? The natural numbers are infinite, so wouldn't the set contain infinitely many elements?

• $[0,1]$ is certainly not finite. Every finite set is compact, but the converse is false: take any infinite set with the indiscrete topology for an easy example. – Forever Mozart Dec 15 '14 at 5:27
• Any finite set is compact. However, the converse is certainly false. – Kaj Hansen Dec 15 '14 at 5:28
• Finite set is a set with finite elements. How many elements does $[0,1]$ have? – Swapnil Tripathi Dec 15 '14 at 5:28
• I am not the downvoter, many users downvote when the question seems very trivial for them. – Swapnil Tripathi Dec 15 '14 at 5:36
• +1 for a question which demonstrates a thought process and outlines previous attempts at an answer. – Neal Dec 15 '14 at 5:55

There are at least two ways you can think of a set being finite: in cardinality or in measure.

In cardinality is just the number of elements in the set. Examples of finite cardinality sets include $\{1\}$, $\{1, 2\}$, $Z_n$, $\{2, 3, 5, 7\}$ and sets with infinitely many items include $\{1, 2, 3, ...\}$, $\{2, 4, 6, ...\}$, $\{1, 1/2, 1/3, ... \}$, $[0,1]$, $(-\infty, \infty)$. The last two examples are not just infinite, they are uncountably infinite.

Now a set having finite measure is a more complicated thing. Basically the measure is the "length" of any set as seen on the number line. I won't really define what it is, but it is intuitive mostly. Just some simple examples of sets with finite measure are $[0,1]$, $[2, 3]$, $[0,1] \cup [2,3]$. Some examples of sets with infinite measure are $(-\infty, \infty)$ and $(0, \infty)$.

Now when we talk about compactness, we are talking about finite sets in cardinality. That's why $[0,1]$ isn't an example of a finite set being compact. It is finite in measure (and also bounded), but it is not finite in cardinality.

No, no, no! Any finite set is compact, but not conversely!

• Haha. That should be a comment!! :D – Swapnil Tripathi Dec 15 '14 at 5:30
• @SwapnilTripathi You are right. But such a question asked by OP with reputations >500 cries for a cry. – Przemysław Scherwentke Dec 15 '14 at 5:32
• @PrzemysławScherwentke for crying out loud!!! haha – Forever Mozart Dec 15 '14 at 5:34
• @SwapnilTripathi: No it shouldn't. It is the answer. – Nick Matteo Dec 15 '14 at 5:51
• What to think of such an answer posted by a user with reputation >4000? – user147263 Dec 15 '14 at 18:55

As already explained by other answers, this is not valid in general, and your example $\mathbb R$ is a space where it is not valid.

But there are actually topological spaces, where the assertion »$A \text{ is compact} \Rightarrow A \text{ is finite}$« is valid.

A trivial example is every topology on a finite set: all subsets of a finite set are finite, thus also all compact subsets of this finite set are finite.

Another example is the discrete topology (on any set), i.e. the topology where every one-element subset is open, or (equivalently) every subset is open.

Proof: Assume $A$ is an infinite set in such a space, then $A = \bigcup_{a \in A} \{ a \}$ is an (infinite) open covering, and we can't find a finite subcovering (as each one-element set is needed). Therefore $A$ is not compact.

(Common examples for discrete topologies are $\mathbb N$ with the topology inherited from $\mathbb R$, or any set with the discrete metric $d(x,y) = \delta_{xy} = \begin{cases}0 & x = y \\ 1 &x \neq y \end{cases}$.)