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Show set of all differentiable functions $f\colon [0,1]\to\mathbb R$ is uncountable.

I think we have to use the Schröder-Bernstein theorem which states that if $|A| \leq |B|$, and $|A| \geq |B|$ then $|A|=|B|$.

Need some help.

Hagen von Eitzen
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josh
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6 Answers6

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Hint: For $c\in\mathbb R$ consider $f(x)=c$.

Hagen von Eitzen
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  • is that a bijection from f to R? – josh Apr 12 '13 at 17:03
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    @josh You did not asked that your set has the same cardinality as $\mathbb{R}$ but that it is uncountalbe hence that it is not countable. – clark Apr 12 '13 at 17:05
  • sorry but i still dont understand why its uncountable, can you explain? uncountable means that the cardinality of the set isnt equal to |N| right? – josh Apr 12 '13 at 17:10
  • @josh Each constant function is differentiable therefore you have the injection $$H: \mathbb{R} \rightarrow \mathrm{Your, set}$$ where one real $c$ is sent to the constant differential function $f(x)=c$ – clark Apr 12 '13 at 17:14
  • @josh No. Uncountable means it is strictly greater than $|\Bbb N|$. $|{1,2,3}|\neq |\Bbb N|$ but they certainly ${1,2,3}$ is countable. – Pedro Apr 12 '13 at 17:15
  • @clark right, so this shows that |R| < set of continuous functions? – josh Apr 12 '13 at 17:15
  • @josh this is right. But it also proves this inequality for the differential functions – clark Apr 12 '13 at 17:18
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To show that this set has the same cardinality as $\mathbb{R}$ proceed as follows. For the one direction notice that your the set of differentiable function is a subsets of the continuous functions so their cardinality is less or equal their cardinality.

A continuous functions is uniquely determined if we know the values it takes at the rationals. Therefore they have the same cardinality as $\mathbb{R}^{\mathbb{Q}}$ but $$\mathbb{R}^{\mathbb{Q}}\cong \mathbb{R}^{\mathbb{N}}\cong {\left (\mathbb{2^\mathbb{N}}\right )}^\mathbb{N} \cong 2^{\mathbb{N} \times \mathbb{N}}\cong2^\mathbb{N} \cong \mathbb{R}. $$ Which proved that the continuous function have less or equal cardinality than $\mathbb{R}$.

clark
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Another way to look at this problem is to use the basics of metric spaces. As we know that continuous image of connected set is connected and $[0,1]$ is connected subset of $R$. Since, differentiable maps from $[0,1]$ to $\mathbb{R}$ must be continuous though. Now, using continuous image of connected sets is connected it is clear that the set of all the continuous maps from $[0,1]$ to $\mathbb{R}$ must be connected subset of $R$ and The only connected subsets of R are the intervals (bounded or unbounded, open orclosed or neither) which must be uncountable.

Srijan
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Take the collection of functions $f_\alpha(x) = \alpha$, with $\alpha \in [0,1]$. Since $[0,1]$ is uncountable, and each $f_\alpha$ is distinct, the set $\{ f_\alpha \}_{\alpha \in [0,1]}$ is uncountable.

Here $A=[0,1]$, $B= \{f:[0,1] \to \mathbb{R} | f \text{ is differentiable} \}$, and $\phi:A \to B$ given by $\phi(\alpha) = f_\alpha$ is injective.

copper.hat
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Let $(f)_r:[0,1]\rightarrow \mathbb{R}$ be a family of functions over $\mathbb{R}$ defined as $f(x)=x^r$. Clearly differentiable, since exponential. Clearly uncountable since defined over $\mathbb{R}$

mathemagician
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In a more general way, using @mathemagician idea, take any one differentiable function over $x$ and make it into a function over $ rx : r \in \mathbb R $ and obtain a continuum worth of differentiable functions through $f(x) \rightarrow f(rx): r \in \mathbb R $ each such $f(r_ox)$ is differentiable for all $r_o \in \mathbb R$ , as the composition of differentiable functions, and there are $\mathbb R=\mathbb c$ such functions.

You can e.g., take $ \sin x \rightarrow sin(rx) : r \in \mathbb R$ . Any one differentiable function will do.

You can see how to use a similar argument to show there are countably-many non-differentiable functions $f: [0,1] \rightarrow \mathbb R$

MSIS
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