real analysis - Why can a closed, bounded interval be uncountable?

From what I have read, all finite sets are countable but not all countable sets are finite. As I understand it,

• Countably Finite --- a one to one map onto $\Bbb{N}$ with a limited number of members


• Countably Infinite --- a one to one map onto $\Bbb{N}$ with an unlimited number of members but that you can count in principle if given an infinite amount of time


• Uncountably Infinite --- there is no one to one mapping onto $\Bbb{N}$. Even if you count, you will miss some of the members. And it is infinite.

From this I gather that countable is not the same as finite. Countable is the one to one property with $\Bbb{N}$. Finite just means a limited number of elements.

Now consider $[0,1]$ which is closed and bounded.

• Bounded --- $\forall k \in [0,1]$ we have $k \leq 1$. Similarly all $k\geq0$.


• Closed --- it contains the endpoints $0$ and $1$

Yet I read $[0,1]$ is uncountably infinite. So clearly, neither closure nor boundedness implies finiteness or countability.

Question:

Why can a closed, bounded interval be uncountable?

It just seems like something that is bounded would be "more finite" than something that isn't.

Answer

I suspect you're conflating two meanings of "finite". Some sets are finite, meaning they have only finitely many elements. An interval like $[0,1]$ is not such a set. On the other hand, $[0,1]$ has finite length, which is a quite different matter. As the other answers have explained, finite length does not imply finiteness (or even countability) in terms of the number of elements.