Unbounded Space Minus Bounded Space is Unbounded/Proof 2

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

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.

$\blacksquare$