# Definition:Bounded Below Set/Real Numbers

*This page is about subsets of the real numbers which are bounded below. For other uses, see Definition:Bounded Below.*

## Contents

## Definition

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

A subset $T \subseteq \R$ is **bounded below (in $\R$)** if and only if $T$ admits a lower bound (in $\R$).

### Unbounded Below

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

## Examples

### 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 $I$.

## Also see

## Sources

- 1964: Walter Rudin:
*Principles of Mathematical Analysis*(2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.33$. Definition - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $\S 1.1$: Real Numbers - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.2$: The Continuum Property - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 10$: The well-ordering principle: Definition $1$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**bound**