Unbounded Space Minus Bounded Space is Unbounded

Theorem

Let $M$ be a metric space.

Let $A \subseteq M$ be unbounded in $M$.

Let $B \subseteq M$ be bounded in $M$.

Then $A \setminus B$ is unbounded in $M$.

Proof 1

Aiming for a contradiction, suppose $A \setminus B$ is bounded.

Then by definition of bounded:

$\exists K \in \R: \forall x, y \in A \setminus B: \map d {x, y} \le K$

By definition of bounded, choose some $a_b \in A$ and $d_b \in \R$ such that:

$\forall x \in B: \map d {x, a_b} \le d_b$

By definition of unbounded, $A$ is not bounded is $M$.

Thus, by the negation of the definition of bounded:

$\exists a_x \in A: \map d {a_x, a_b} > d_b$

Let $d_x = \map d {a_x, a_b}$.

Again, by the negation of the definition of bounded:

$\exists a_y \in A: \map d {a_y, a_b} > d_x + K$

Let $d_y = \map d {a_y, a_b}$.

As $\map d {a_x, a_b} > d_b$, it follows that $a_x \notin B$.

Likewise, as $\map d {a_y, a_b} > d_x > d_b$, $a_y \notin B$.

Thus, both $a_x, a_y \in A \setminus B$.

But:

\(\ds \map d {a_y, a_x}\)

\(\ge\)

\(\ds \size {\map d {a_b, a_y} - \map d {a_b, a_x} }\)

Reverse Triangle Inequality

\(\ds \)

\(=\)

\(\ds \size {d_y - d_x}\)

\(\ds \)

\(=\)

\(\ds d_y - d_x\)

As $d_y > d_x$

\(\ds \)

\(>\)

\(\ds d_x + K - d_x\)

\(\ds \)

\(=\)

\(\ds K\)

Then, $a_y, a_x \in A \setminus B$ are two points such that $\map d {a_y, a_x} > K$, contradicting our assumption.

Thus, by Proof by Contradiction, it follows that $A \setminus B$ is unbounded.

$\blacksquare$

Proof 2

Aiming for a contradiction, suppose $A \setminus B$ is bounded.

By Finite Union of Bounded Subsets:

$\paren {A \setminus B} \cup B$

is bounded.

On the other hand, by Definition of Set Difference:

$A \subseteq \paren {A \setminus B} \cup B$

Thus by Subset of Bounded Subset of Metric Space is Bounded, $A$ must be bounded, too.

This is a contradiction.

$\blacksquare$