Prove that the set of all differentiable functions :[0,1]→ [0,1] is uncountable.

Question

Prove that the set of all differentiable functions :[0,1]→ [0,1] is uncountable.

In my notes, I have something like:

Consider $x \in {[0,1]}$, $f_{x} (t) = x$ for all $t$ . $\{h_{x} | x ∈ [0,1] \}= |[0,1]| = c$ .

I have no idea what this means. Can anyone explain or help me prove it?

Answer 1

For each $x \in [0,1]$ define $f_x(t)=x$ for all $t \in [0,1]$. This gives you an element $f_x$ of your set since constant functions are differentiable. The map $x \to f_x$ is one to one ($f_x=f_y$ implies $f_x(0)=f_y(0)$ which means $x=y$) and $[0,1]$ is uncountable. Hence the given set is also uncountable.

Answer 2

Actually, some concerns... It is not for all $t$ but for $t\in{[0, 1]}$. Second for the set described in your question, it assumes the constant functions describes in the comments. Third, the $c$ in your question is the continoum symbol.

Answer 3

To clarify the other answers:

• Each value in $[0,1]$ defines a distinct constant function
• Each of these constant functions is differentiable
• The collection of these functions has the same cardinality as $[0,1]$ since the functions are all distinct
• The cardinality of $[0,1]$ is the uncountable cardinal $\mathfrak c$