| arc length |
$1$
|
The arc length of $[0, 1]$, as a straight segment, is its length, $1 - 0 = 1$.
|
| diameter |
$1$
|
The diameter of $[0, 1]$ is $1$, the distance between its endpoints.
|
| cardinality |
$\mathfrak{c}$
|
The cardinality of $[0, 1]$ is $\mathfrak{c}$, the same as that of $\reals$, because Cantor’s diagonal argument shows that $[0, 1]$ is uncountable, and $[0, 1]$ is in bijection with $\reals$.
|
| measure |
$1$
|
The Lebesgue measure of $[0, 1]$ is its length, $1$.
|
| Hausdorff dimension |
$1$
|
The Hausdorff dimension of $[0, 1]$ is $1$, as for every segment.
|
| interior |
$\operatorname{int} [0, 1] = (0, 1)$
|
The interior of $[0, 1]$ is $(0, 1)$, because no interval around $0$ or $1$ stays inside $[0, 1]$.
|
| boundary |
$\partial [0, 1] = \setdef{0, 1}$
|
The boundary of $[0, 1]$ is $\setdef{0, 1}$, because every interval around $0$ or $1$ contains points inside and outside $[0, 1]$.
|
| exterior |
$\operatorname{ext} [0, 1] = (-\infty, 0) \cup (1, \infty)$
|
The exterior of $[0, 1]$ is everything outside $[0, 1]$, because each of those points has a small interval around it that misses $[0, 1]$.
|
| closure |
$\overline{[0, 1]} = [0, 1]$
|
The closure of $[0, 1]$ is $[0, 1]$ itself, because $[0, 1]$ is closed.
|
| closed set |
✓ |
The interval $[0, 1]$ is closed in $\reals$, because it contains its boundary points $0$ and $1$.
|
| compact |
✓ |
$[0, 1]$ is compact because it is closed and bounded in $\reals$.
|
| sequentially compact |
✓ |
$[0, 1]$ is sequentially compact (as a compact metric space), locally compact, and paracompact, because it is compact.
|
| locally compact |
✓ |
|
| paracompact |
✓ |
|
| bounded |
✓ |
$[0, 1]$ is compact because it is closed and bounded.
|
| infinite |
✓ |
$[0, 1]$ is uncountable, and so infinite, because Cantor’s diagonal argument shows it cannot be listed.
|
| uncountable |
✓ |
$[0, 1]$ is uncountable because Cantor’s diagonal argument shows that no list of numbers in $[0, 1]$ contains them all.
|
| open set |
✗ |
The interval $[0, 1]$ is not open in $\reals$, because every interval around $0$ contains negative numbers, which are not in $[0, 1]$.
|
| finite |
✗ |
$[0, 1]$ is not finite, because it has infinitely many elements.
|
| countably infinite |
✗ |
$[0, 1]$ is not countably infinite, because it is uncountable.
|
| countable |
✗ |
$[0, 1]$ is not countable, because it is uncountable.
|