1.2
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Prove that \((a, b)\) is uncountable \(\forall a < b\).
Since \(\mathbb{R} \rightarrow \mathbb{R}_{>0} \rightarrow (0, 1)\)Uncountable, then i will prove that \(\exists f : (a, b) \to (0, 1)\) bijection.
So \((a, b)\) will have to be uncountable because in lecture we proved that \((0, 1)\) is uncountable.We can construct a linear function that
After calculating this, we get \(f(x)=\frac{x-a}{b-a}\)
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Let \(A, B\) be two sets such that \(A \subseteq B\). Prove that if \(A\) is countable and \(B\) is uncountable, then \(B - A\) is uncountable.
Contradiction, Suppose \(B-A\) is countable, then \((B-A)\cup (A)=B\) is also countable.
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Let \(A, B\) be two sets such that \(A \subseteq B\). Prove that if \(A\) is uncountable, then \(B\) is uncountable.
Contradiction, Suppose \(B\) is countable, then since \(A\subseteq B\), then \(A\) is also countable.
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Prove that the following set is denumerable: \(A = \{(x, y) \in \mathbb{N} \times \mathbb{R} \mid x y = 1\}\)
I will prove the existence of a bijection \(f: A \to \mathbb{N}\). And so I will have proved that \(A\) is denumerable by definition \(f:A\to \N\) where \((x,y)\mapsto x\)
Let's see that \(f\) is injective and surjective
Injectivity If \(f(x, z) = f(x, y) = x\). Since \((x, y), (x, z) \in A\), we have \(x z = 1\) and \(x y = 1\). Then \(z = \frac{1}{x}\), \(y = \frac{1}{x}\) \(\Rightarrow\) \((x, y) = (x, z)\).
Surjectivity Let \(n \in \mathbb{N}\). We need to show that \(\exists (x, y) \in A\) such that \(f(x, y) = n\). Choose \((x,y)=(n,\frac1n)\)
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Prove that there exists a bijection from these sets to \(\mathbb{R}\): a) \(\mathbb{R} - \{0\}\)
\(f: \mathbb{R} \to \mathbb{R} - \{0\}\), bijection \(x \mapsto x\) if \(x \notin \mathbb{N} \cup \{0\}\) \(x \mapsto x + 1\) if \(x \in \mathbb{N} \cup \{0\}\)
Injective: If \(f(x) = f(y)\), three cases
Surjective: Let \(r \in \mathbb{R} - \{0\}\). I will prove that \(\exists x \in \mathbb{R}\) such that \(f(x) = r\).
- \(x = r - 1\) if \(r \in \mathbb{N}\)
- \(x = r\) if \(r \notin \mathbb{N}\)
(b) using a draw