Definition:Bounded Above Set/Real Numbers

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This page is about Bounded Above of Subset of Real Numbers. For other uses, see Bounded Above.

Definition

Let $\R$ be the set of real numbers.

A subset $T \subseteq \R$ is bounded above (in $\R$) if and only if $T$ admits an upper bound (in $\R$).


Unbounded Above

$T \subseteq \R$ is unbounded above (in $\R$) if and only if it is not bounded above.


Examples

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 $I$.


Also see

  • Results about bounded above sets of real numbers can be found here.


Sources

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