The Sup. and Inf. Properties for Subsets of Sets of Real Numbers

The Supremum and Infimum Properties for Subsets of Sets of Real Numbers

If $A$ is a nonempty set of real numbers that is bounded above (or below) then we know that $\sup A$ (or $\inf A$) exists. When $B \subseteq A$ is also nonempty we can relate $\sup A$ to $\sup B$ (or $\inf A$ to $\inf B$). The following results show the relation

Theorem 1: Let $A$ and $B$ be nonempty subsets of real numbers such that $B \subseteq A$. If $A$ is bounded above then $\sup B \leq \sup A$.

• Proof: Let $A, B \subseteq \mathbb{R}$ be nonempty with $B \subseteq A$ and let $A$ be bounded above. Since $A$ is bounded above, every subset of $A$ is also bounded above. Since $B \subseteq A$ this means that $B$ is a nonempty subset of real numbers that is bounded above. Thus $\sup B$ is well-defined.

• Suppose that $\sup B > \sup A$. Then $M = \sup A$ is not an upper bound for $B$. So there exists an element $x \in B$ such that $M < x$. But $B \subseteq A$ so $x \in A$. So there exists an element $x \in A$ such that $M < x$. But this is a contradiction since $M$ is an upper bound of $A$. Therefore:

(1)

\begin{align} \quad \sup B \leq \sup A \quad \blacksquare \end{align}

Corollary 2: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers that is bounded above. Then $\displaystyle{\left ( \sup_{k \geq n} \{ a_k \} \right )_{n=1}^{\infty}}$ is a decreasing sequence.

• Proof: We note that since $(a_n)_{n=1}^{\infty}$ is bounded above then the set $\{ a_k : k \geq n \}$ is bounded above for every $n \in \mathbb{N}$. Also:

(2)

\begin{align} \quad \{ a_k : k \geq 1 \} \supseteq \{ a_k : k \geq 2 \} \supseteq ... \end{align}

• By the previous theorem this means that:

(3)

\begin{align} \quad \sup \{ a_k : k \geq 1 \} \geq \sup \{ a_k : k \geq 2\} \geq ... \end{align}

• From the inequality above, $\displaystyle{(\sup_{k \geq n} \{ a_k \})_{n=1}^{\infty}}$ is a decreasing sequence. $\blacksquare$

Theorem 3: Let $A$ and $B$ be nonempty subsets of real numbers such that $B \subseteq A$. If $A$ is bounded below then $\inf B \geq \inf A$.

• Proof: Let $A, B \subseteq \mathbb{R}$ be nonempty with $B \subseteq A$ and let $A$ be bounded below. Since $A$ is bounded below, every subset of $A$ is also bounded below. Since $B \subseteq A$ this means that $B$ is a nonempty subset of real numbers that is bounded below. Thus $\inf B$ is well-defined.

• Suppose that $\inf B < \inf A$. Then $m = \inf A$ is not a lower bound for $B$. So there exists an element $x \in B$ such that $x < m$. But $B \subseteq A$ so $x \in A$. So there exists an element $x \in A$ such that $x < m$. But this is a contradiction since $m$ is a lower bound of $A$. Therefore:

(4)

\begin{align} \quad \inf B \geq \inf A \quad \blacksquare \end{align}

Corollary 4: Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers that is bounded below. Then $\displaystyle{\left ( \inf_{k \geq n} \{ a_k \} \right )_{n=1}^{\infty}}$ is an increasing sequence.

• Proof: We note that since $(a_n)_{n=1}^{\infty}$ is bounded below then the set $\{ a_k : k \geq n \}$ is bounded below for every $n \in \mathbb{N}$. Also:

(5)

\begin{align} \quad \{ a_k : k \geq 1 \} \supseteq \{ a_k : k \geq 2 \} \supseteq ... \end{align}

• By the previous theorem this means that:

(6)

\begin{align} \quad \sup \{ a_k : k \geq 1 \} \leq \sup \{ a_k : k \geq 2\} \leq ... \end{align}

• From the inequality above, $\displaystyle{(\sup_{k \geq n} \{ a_k \})_{n=1}^{\infty}}$ is an increasing sequence. $\blacksquare$