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Cardinality of continuous functions $f:\Bbb R\to \Bbb R$.

We know that a continuous function $f:\Bbb R\to \Bbb R$ is completely determined by its values on $\Bbb Q$.

For each rational $q\in \Bbb Q,f(q)$ has $c(=\text{Card }\Bbb R)$ many choices.

Hence the number of such functions becomes $\mathfrak{q^c}= \Bbb N^c$ .

Is it right?Please check.

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It is not correct. For starters, the expression $q^\mathfrak{c}$ (where $q\in\Bbb Q$) is meaningless; I suspect that you meant $\left|\Bbb Q^\mathfrak{c}\right|$, which is indeed $\left|\Bbb N^\mathfrak{c}\right|$. But note that you do need the cardinality bars.

However, you’ve got the multiplication principle backwards: you’re making a choice amongst $\mathfrak{c}$ real numbers $|\Bbb Q|$ times, so you really want $\mathfrak{c}^{|\Bbb N|}$. Finally, $\mathfrak{c}=2^{|\Bbb N|}$, so

$$\mathfrak{c}^{|\Bbb N|}=\left(2^{|\Bbb N|}\right)^{|\Bbb N|}=2^{|\Bbb N|\cdot|\Bbb N|}=2^{|\Bbb N|}=\mathfrak{c}\;.$$

This actually shows only that there are at most $\mathfrak{c}$ continuous real-valued functions on $\Bbb R$, since it counts all functions from $\Bbb Q$ to $\Bbb R$, not just the ones that are restrictions of continuous functions from $\Bbb R$ to $\Bbb R$. To complete the argument you need to show that there are also at least that many; this is easy, since there are $\mathfrak{c}$ constant functions.

I’ve answered this in order to respond to the specific problems with your argument. The question itself has been dealt with in detail in the answers to this question.

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Brian M. Scott
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  • One question?Why can't we conclude from the first para that there are exactly $\mathfrak c$ functions?Why do we state at most $\mathfrak c$ many functions?Since for each $q\in \Bbb Q\exists\mathfrak c$ many choices so can't we conclude that number of such functions is exactly $\mathfrak c$ as above – Learnmore Dec 14 '16 at 03:02
  • Why do we need to use constant functions to guarantee at least $\mathfrak c$?Would be glad if u could help. – Learnmore Dec 14 '16 at 03:03
  • @Ben: You’re welcome. \ The functions that you’re counting aren’t necessarily continuous; there are $\mathfrak{c}$ of them, but it’s conceivable that fewer than that many are continuous. \ We don’t have to use the constant functions to guarantee that there are at least $\mathfrak{c}$ continuous functions; any set of $\mathfrak{c}$ continuous functions would work. But they’re one of the easiest ways to do so: they are very easily seen to be continuous, and it’s immediately clear that there are $\mathfrak{c}$ of them. – Brian M. Scott Dec 14 '16 at 17:23
  • @Ben: You’re very welcome. – Brian M. Scott Dec 14 '16 at 17:46