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How to prove that for any function $f: \mathbb{R} \rightarrow \mathbb{R}$ there exist two injections $g,h \in \mathbb{R}^{\mathbb{R}} \ : \ g+h=f$.

Could you help me?

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    $\begingroup$ Are you familiar with transfinite induction? $\endgroup$ Commented Jan 14, 2013 at 21:00

1 Answer 1

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Note that it suffices to find an injection $g$ such that $f+g$ is also an injection, as $f$ can then be written as the sum $(f+g)+(-g)$. Such a $g$ can be constructed by transfinite recursion.

Let $\{x_\xi:\xi<2^\omega\}$ be an enumeration of $\Bbb R$. Suppose that $\eta<2^\omega$, and we’ve defined $g(x_\xi)$ for all $\xi<\eta$ in such a way that $g(x_\xi)\ne g(x_\zeta)$ and $(f+g)(x_\xi)\ne(f+g)(x_\zeta)$ whenever $\xi<\zeta<\eta$. Let

$$S_\eta=\{g(x_\xi):\xi<\eta\}$$

and

$$T_\eta=\{(f+g)(x_\xi)-f(x_\eta):\xi<\eta\}\;.$$

Then $|S_\eta\cup T_\eta|<2^\omega$, so we may choose $g(x_\eta)\in\Bbb R\setminus(S_\eta\cup T_\eta)$. Clearly $g(x_\eta)\ne g(x_\xi)$ for $\xi<\eta$, since $g(x_\eta)\notin S_\eta$. Moreover, $g(x_\eta)\notin T_\eta$, so for each $\xi<\eta$ we have $g(x_\eta)\ne(f+g)(x_\xi)-f(x_\eta)$ and hence $(f+g)(x_\eta)\ne(f+g)(x_\xi)$. Thus, the recursion goes through to define $g(x_\xi)$ for all $\xi<2^\omega$, and it’s clear from the construction that both $g$ and $f+g$ are injective.

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  • $\begingroup$ is it possible to prove if we do not assume that $\mathbb{R}$ can be well-ordered? $\endgroup$
    – Ali Dursun
    Commented Oct 6, 2021 at 13:57
  • $\begingroup$ @AliDursun: I don’t know. If I had to guess, it guess not, but I don’t actually know of a model in which it fails. That would be a good question for Asaf Karagila. $\endgroup$ Commented Oct 6, 2021 at 23:50

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