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Let $F=$ $\{$ $f: [0,1] \rightarrow \mathbb{R}$ $:$ $f$ is constant$ \} $. I must show that $F$ is uncountable. Note, that for any $f \in F$, and any $c\in \mathbb{R}$, I will denote the constant function $f:$ [0,1] $\rightarrow \mathbb{R}$, $f(x)=c$ by $f_c$ .

Define a map $g: \mathbb{R} \rightarrow F$ by $g(x)= f_x$ then observe that $g(a)=g(b)$ $\implies$ $f_a=f_b$. $f_a$ is the map defined on $[0,1]$ given by $f(x)=a$ and $f_b$ is the map defined on $[0,1]$ given by $h(x)=$ $b$. As the maps are equal, $a=b$ and so $g$ is injective, so $F$ must be uncountable.

Note: The map is also surjection and consequently a bijection, but showing it is indeed a bijection is not necessary.

Is the proof correct? What other easier ways are there to prove this?

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  • $\begingroup$ Why was this downvoted? $\endgroup$
    – user643073
    Commented Oct 30, 2019 at 11:42
  • $\begingroup$ Looks fine to me albeit a bit clunky. $\endgroup$
    – Cameron Williams
    Commented Oct 30, 2019 at 11:57
  • $\begingroup$ May I see how you write it? Just so I could try to make my proofs look more elegant, please? $\endgroup$
    – user643073
    Commented Oct 30, 2019 at 11:58
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    $\begingroup$ Typo: You say $g:\mathbb R \to \mathbb R$ but it should be $g:\mathbb R \to F$ $\endgroup$
    – lulu
    Commented Oct 30, 2019 at 12:08
  • $\begingroup$ @lulu stole my thunder haha $\endgroup$
    – Cameron Williams
    Commented Oct 30, 2019 at 12:10

2 Answers 2

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It would have been simpler to define a bijection $g:\,F\mapsto\Bbb R$ by $g(f):=f(0)$ for all $f\in F$.

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Follow up problem.
Let S be a subset of R.
Show there are |S| many constant functions from [0,1] into S.

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