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 $I$.