Answer
See explanation
Work Step by Step
To show that the set of all odd integers,
\[
\mathcal{O} \;=\; \{\, 2k + 1 \mid k \in \mathbb{Z} \},
\]
has the same cardinality as the set of all even integers,
\[
2\mathbb{Z} \;=\; \{\, 2k \mid k \in \mathbb{Z} \},
\]
it suffices to exhibit a bijection between them. One simple choice is:
\[
f : \mathcal{O} \;\to\; 2\mathbb{Z},
\quad f(2k + 1) \;=\; 2k.
\]
---
### 1. Injectivity
Suppose \(f(2k + 1) = f(2m + 1)\). Then
\[
2k \;=\; 2m
\quad\Longrightarrow\quad
k \;=\; m
\quad\Longrightarrow\quad
2k + 1 \;=\; 2m + 1.
\]
Hence, if two odd integers map to the same even integer, they must be the same odd integer. Thus, \(f\) is injective.
---
### 2. Surjectivity
Every even integer can be written as \(2n\) for some \(n \in \mathbb{Z}\). Observe that
\[
f(2n + 1) \;=\; 2n,
\]
so for any even integer \(2n \in 2\mathbb{Z}\), there is an odd integer \(2n + 1 \in \mathcal{O}\) that maps to it. Hence, \(f\) is surjective.
---
Because \(f\) is both injective and surjective, it is a **bijection**, and therefore \(\mathcal{O}\) (the set of odd integers) and \(2\mathbb{Z}\) (the set of even integers) have the same cardinality.
