Bounded Below Subset of Real Numbers/Examples

From ProofWiki
Jump to navigation Jump to search

Examples of Bounded Below Subsets of Real Numbers

Example: $\openint 1 \to$

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

$S = \openint 1 \to$

is bounded below.


Examples of lower bounds of $S$ are:

$-7, 1, \dfrac 1 2$


The set of all lower bounds of $S$ is:

$\hointl {-\infty} 1$


Example: $\set {x \in \R: x > 0}$

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

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

is bounded below, but unbounded above.


Let $H > 0$ in $T$ be proposed as an upper bound.

Then it is seen that $H + 1 \in T$ and so $H$ is not an upper bound at all.


Examples of lower bounds of $T$ are:

$-27, 0$

Its infimum is $0$.


Example: $\hointl {-\infty} 2$

Let $I$ be the open real interval defined as:

$I := \openint 0 1$

Then $I$ is not bounded below.

Hence $I$ does not admit an infimum, and so does not have a smallest element.


Example: $\openint 0 1$

Let $I$ be the open real interval defined as:

$I := \openint 0 1$

Then $I$ is bounded below by, for example, $0$, $-1$ and $-2$, of which $0$ is the infimum.

However, $I$ does not have a smallest element.


Example: $\set {-1, 0, 2, 5}$

Let $I$ be the set defined as:

$I := \set {-1, 0, 2, 5}$

Then $I$ is bounded below by, for example, $-1$, $-2$ and $3$, of which the infimum is $-1$.

$5$ is also the smallest element of $I$.


Example: $\openint 3 \infty$

Let $I$ be the unbounded open real interval defined as:

$I := \openint 3 \to$

Then $I$ is bounded below by, for example, $3$, $2$ and $1$, of which the infimum is $3$.

However, $I$ does not have a smallest element.


Example: $\closedint 0 1$

Let $I$ be the closed real interval defined as:

$I := \closedint 0 1$

Then $I$ is bounded below by, for example, $0$, $-1$ and $-2$, of which $0$ is the infimum.

$I$ is also the smallest element of $I$.


Example: $\openint \gets \to$

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

$\R$ is not bounded below.


Example: $\openint \gets 0$

Let $I$ be the unbounded open real interval defined as:

$I := \openint \gets 0$

Then $I$ is unbounded below.


Example: $\hointl 0 1$

Let $I$ be the left half-open real interval defined as:

$I := \hointl 0 1$

Then $I$ is bounded below.

$0$ is an lower bound of $I$.

Retrieved from "https://proofwiki.org/w/index.php?title=Bounded_Below_Subset_of_Real_Numbers/Examples&oldid=652104"