Answer
Because \(f\) is both one‐to‐one and onto, it is a **bijection**, showing that \((0,1)\) and \((0,2)\) have the **same cardinality** (the cardinality of the continuum).
Work Step by Step
To prove that the open interval \(S = (0,1)\) has the same cardinality as \(U = (0,2)\), it suffices to exhibit a **bijection** between them. A simple choice is:
\[
f: (0,1) \,\to\, (0,2),
\quad f(x) = 2x.
\]
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### 1. \(f\) is One‐to‐One (Injective)
If \(f(x_1) = f(x_2)\), then \(2x_1 = 2x_2\). Dividing both sides by 2 yields \(x_1 = x_2\). Hence \(f\) is injective.
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### 2. \(f\) is Onto (Surjective)
Any point \(y \in (0,2)\) can be written as \(y = 2x\) for some \(x \in (0,1)\). Specifically, take \(x = \tfrac{y}{2}\). Since \(y\) is between 0 and 2, \(\tfrac{y}{2}\) is between 0 and 1, which lies in the domain. Thus every point in \((0,2)\) is hit by some \(x \in (0,1)\), so \(f\) is surjective.
