Supremum of Subset of Real Numbers/Examples

From ProofWiki
Jump to navigation Jump to search

Examples of Suprema of Subsets of Real Numbers

Example 1

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {1, 2, 3}$

admits a supremum:

$\sup S = 3$


Example 2

The subset $T$ of the real numbers $\R$ defined as:

$T = \set {x \in \R: 1 \le x \le 2}$

admits a supremum:

$\sup T = 2$


Example 3

The subset $V$ of the real numbers $\R$ defined as:

$V := \set {x \in \R: x > 0}$

does not admit a supremum.


Example 4

Consider the set $A$ defined as:

$A = \set {3, 4}$

Then the supremum of $A$ is $4$.

However, $A$ contains no element $x$ such that:

$3 < x < 4$.


Example: $\openint \gets 0$

Let $\R_{<0}$ be the (strictly) negative real numbers:

$\R_{<0} := \openint \gets 0$

Then the supremum of $\R_{<0}$ is $0$.


Example: $\hointl 0 1$

Let $\hointl 0 1$ denote the left half-open real interval:

$\hointl 0 1 := \set {x \in \R: 0 < x \le 1}$

Then the supremum of $\R_{<0}$ is $1$.


Example 5

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {x \in \R: x^2 \le 2 x - 1}$

admits a supremum:

$\sup S = 1$

such that $\sup S \in S$.


Example 6

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {x \in \R: x^2 + 2 x \le 1}$

admits a supremum:

$\sup S = -1 + \sqrt 2$

such that $\sup S \in S$.


Example 7

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {x \in \R: x^3 < 8}$

admits a supremum:

$\sup S = 2$

such that $\sup S \notin S$.


Example 8

The subset $S$ of the real numbers $\R$ defined as:

$S = \set {x \in \R: x \sin x < 1}$

does not admit a supremum.