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We need to show that the set of real numbers and the set of Real valued functions whose domain is $\mathbb R$ are not similar (equinumerous).

Let $\mathbb R$ denote the set of real numbers and $S$ denote the set of real valued functions whose domain is $\mathbb R$.

Suppose $\mathbb R$ and $S$ are equinumerous. Then, $~\exists ~$ a one-one onto function $f :\mathbb R \rightarrow S $ such that $f( \mathbb R) = S$.

Let $a \in \mathbb R$, then let $g_a$ be the associated real valued function with $a$. Thus :

$f(a) = g_a$.

To bring a contradiction, I think we should show that $f$ is either not one-one or onto.

How do I move forward?

Thank you for your help.

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    $\begingroup$ Trying to show that $f$ is not one-one is not a promising approach because there are at least as many real functions as there are numbers. For instance, what if $g_a$ is just the constant function $a$? So the way forward is to try and show that $f$ is not onto, which means, find a real function which is different from every $g_a$. $\endgroup$
    – bof
    Sep 28, 2014 at 4:43
  • $\begingroup$ Suppose $f(a) = g_a$ , then $1 + g_a = f(b)$ for some $b \in \mathbb R \implies 1 + f(a) = f(b) \implies f(b) - f(a) = 1$. Am I on the correct path? $\endgroup$
    – MathMan
    Sep 28, 2014 at 5:16
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    $\begingroup$ I doubt it. To show that $f$ is not onto, you have to construct a function $h:\mathbb R\to\mathbb R$ such that $h\ne g_a$ for each $a\in\mathbb R$. That means that, for each $a$, there is some $x$ such that $h(x)\ne g_a(x)$. The simplest way is to take $x=a$, that is, define a function $h$ so that $h(a)\ne g_a(a)$ for each $a$. For instance, $h(a)=g_a(a)+17$. $\endgroup$
    – bof
    Sep 28, 2014 at 6:30
  • $\begingroup$ I get it now. Thank you very much :-) $\endgroup$
    – MathMan
    Sep 28, 2014 at 9:06

3 Answers 3

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Don't move forward. As often happens, you need to move sideways, backwards, different sideways, forward, sideways, sideways, backwards, sideways, forward in order to actually move forward.

Note that the set of functions from $\Bbb R$ into $\{0,1\}$ is a subset of all real-valued functions, and remember Cantor's theorem.

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    $\begingroup$ (In general, the best way to move is by doing two quarter-circles forward, and then all three punch keys. HADOUKEN!) $\endgroup$
    – Asaf Karagila ♦
    Sep 28, 2014 at 4:49
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Suppose $F$ is a function from $\mathbb{R}$ to $\mathbb{R}^{\mathbb{R}}$ (the latter is the set of real-valued functions on the reals). We show that $F$ is not onto:

For every $x \in \mathbb{R}$, define $f(x) = F(x)(x)+1$. This well-defined, because $F(x)$ is some function on $\mathbb{R}$ that we can evaluate at $x$, we get a real number and add 1 to it.

Suppose that there exists some $p \in \mathbb{R}$ such that $F(p) = f$. Then $f(p) = F(p)(p) + 1$ (by definition of $f$) $ = f(p) + 1$ as $F(p) = f$. So $f(p) = f(p) +1$, which cannot be (or we'd have $0 = 1$). So no such $p$ can exist. So our $f$ shows that $F$ is not surjective.

As the map that sends each $x \in \mathbb{R}$ to the constant function $f_x$ with value $x$ is clearly injective, we have an injection from $\mathbb{R}$ into $\mathbb{R}^{\mathbb{R}}$ but we cannot have a bijection, so $|\mathbb{R}| < |\mathbb{R}^{\mathbb{R}}|$.

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It should be obvious that an injection from $\mathbb R$ to S is possible, given by assigning each real number $r$ to the constant function $g(x)=r$. In fact, we have a bijection between $\mathbb R$ and the set of all constant functions. Now your problem is to show that there is strictly 'more' functions than constant functions. Assume you have a bijection, and work from there for a contradiction

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  • $\begingroup$ Oh okay .. So, since, there exist a lot more functions than the constant function $g_a(x)=a$ such as $g(x) = ax$, then the mapping $f$ should not be onto. Am I correct .. Is that it? $\endgroup$
    – MathMan
    Sep 28, 2014 at 4:52
  • $\begingroup$ Correct, but that "should be" needs to be proved, as often times in set theory things that you think "should be" more actually aren't. $\endgroup$
    – Alan
    Sep 28, 2014 at 4:53
  • $\begingroup$ Okay .. Our aim is to prove that there are a lot more functions than the set of real numbers.Let $f : \mathbb R \rightarrow S$ be a bijection. I am not able to formally write down what I am thinking .. uhmm, since, there are many more functions other than constant functions , so ... ? Could you give me a hint? $\endgroup$
    – MathMan
    Sep 28, 2014 at 5:04

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