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Math Help - Tricky Function Question

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    MHF Contributor Drexel28's Avatar
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    Tricky Function Question

    Problem: Prove there is some f:\mathbb{R}\to\mathbb{Q} such that a) f(x+y)=f(x)+f(y), b) f\mid_{\mathbb{Q}}=\text{id}_{\mathbb{Q}}, and c) f is not continuous on \mathbb{R}
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    Quote Originally Posted by Drexel28 View Post
    Problem: Prove there is some f:\mathbb{R}\to\mathbb{Q} such that a) f(x+y)=f(x)+f(y), b) f\mid_{\mathbb{Q}}=\text{id}_{\mathbb{Q}}, and c) f is not continuous on \mathbb{R}
    let \{a_i \}_{i \in I}, where a_k=1 for some k \in I, be a basis for \mathbb{R}, as a vector space over \mathbb{Q}. that means every element of \mathbb{R} can be written uniquely as \sum_{i \in I} r_ia_i, where r_i \in \mathbb{Q} and all but at most finitely
    many of r_i are zero. define f: \mathbb{R} \longrightarrow \mathbb{Q} by f \left(\sum_{i \in I} r_ia_i \right)=r_k. the map f is clearly well-defined and satisfies the conditions a) and b). to show that it also satisfies the condition c), choose i \in I
    such that i \neq k and also choose a sequence \{s_n \} \subset \mathbb{Q} such that \lim_{n \to \infty} s_n = \frac{1}{a_i}. now define x_n=a_k - s_na_i and see that \lim_{n \to \infty}f(x_n)=1 but f( \lim_{n\to\infty} x_n)=0.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    let \{a_i \}_{i \in I}, where a_k=1 for some k \in I, be a basis for \mathbb{R}, as a vector space over \mathbb{Q}. that means every element of \mathbb{R} can be written uniquely as \sum_{i \in I} r_ia_i, where r_i \in \mathbb{Q} and all but at most finitely
    many of r_i are zero. define f: \mathbb{R} \longrightarrow \mathbb{Q} by f \left(\sum_{i \in I} r_ia_i \right)=r_k. the map f is clearly well-defined and satisfies the conditions a) and b). to show that it also satisfies the condition c), choose i \in I
    such that i \neq k and also choose a sequence \{s_n \} \subset \mathbb{Q} such that \lim_{n \to \infty} s_n = \frac{1}{a_i}. now define x_n=a_k - s_na_i and see that \lim_{n \to \infty}f(x_n)=1 but f( \lim_{n\to\infty} x_n)=0.
    Good. I had a similar construction, except I kind of drew mine out a little more. It all depended in choosing a Hamel basis.
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