Bounded Above Subset of Real Numbers/Examples
Examples of Bounded Above Subsets of Real Numbers
Example: $\hointl \gets 2$
The subset $I$ of the real numbers $\R$ defined as:
- $I = \hointl \gets 2$
is bounded above by, for example, $2$, $3$ and $4$, of which the supremum is $2$.
$2$ is also the greatest element of $I$.
The set of all upper bounds of $I$ is:
- $\closedint 2 \to$
Example: $\openint 0 1$
Let $I$ be the open real interval defined as:
- $I := \openint 0 1$
Then $I$ is bounded above by, for example, $1$, $2$ and $3$, of which $1$ is the supremum.
However, $I$ does not have a greatest element.
Example: $\set {-1, 0, 2, 5}$
Let $I$ be the set defined as:
- $I := \set {-1, 0, 2, 5}$
Then $I$ is bounded above by, for example, $5$, $6$ and $7$, of which the supremum is $5$.
$5$ is also the greatest element of $I$.
Example: $\openint 3 \to$
Let $I$ be the unbounded open real interval defined as:
- $I := \openint 3 \to$
Then $I$ is not bounded above.
Hence $I$ does not admit a supremum, and so does not have a greatest element.
Example: $\closedint 0 1$
Let $I$ be the closed real interval defined as:
- $I := \closedint 0 1$
Then $I$ is bounded above by, for example, $1$, $2$ and $3$, of which $1$ is the supremum.
$I$ is also the greatest element of $I$.
Example: $\openint \gets \to$
Let $\R$ denote the set of real numbers.
$\R$ is not bounded above.
Example: $\openint \gets 0$
Let $I$ be the unbounded open real interval defined as:
- $I := \openint \gets 0$
Then $I$ is bounded above.
$0$ is an upper bound of $I$.
Example: $\hointl 0 1$
Let $I$ be the left half-open real interval defined as:
- $I := \hointl 0 1$
Then $I$ is bounded above.
$1$ is an upper bound of $I$.