Consider any $a,b\in\mathbb{R}$ such that $a<b$.
Question: Is $[a,b]$ a finite set or not?
The question popped up in my mind while I was working with the Extreme Value Theorem of continuous functions. Clearly, the set has a lower bound and an upper bound, but the set itself can have infinite elements, as you can always find a real number between any two real numbers.
The question can be stated in another way too: Is a set finite if it has finite number of elements, or is a set finite if it has well defined upper and lower bounds?
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$\begingroup$A set is finite (per definition) if it contains only finitely many elements. The interval (more like: the set) $[a,b]$ is infinite because it contains infinitely many elements and it is bounded because it has both lower and upper bounds.
$\endgroup$ $\begingroup$According to set theory (and in particular to the theory of cardinalities), a set $A$ is infinite if there exists a bijection between $A$ and a proper subset of $A$. For any closed interval $[a,b]$ it is very easy to establish this. For any $\varepsilon \in (a,b)$ where $2\varepsilon < b$, we can take
$$f(x) = (x-a) \cdot \frac{\varepsilon}{b-a} + \varepsilon$$
This $f$ (whose graph is easily seen to be a line segment) is a bijection between $[a,b]$ and $[\varepsilon,2\varepsilon]$.
$\endgroup$ $\begingroup$It's infinite, but what are you thinking is about of the measure of the interval, the measure of the interval is finite but the interval itself as a set not.
$\endgroup$ $\begingroup$In short, we know that $(0,1)$ is infinite set, thus $(a,b)$ is infinite.
Let's look at $[a,b]$. We see that $(a,b)$ is a subset of $[a,b]$, thus $[a,b]$ is infinite.
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