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I know that the set of continuous functions on $\mathbb{R^n}$ has cardinality of $\mathbb{R}$. Can this be generalized to any subspace of $\mathbb{R^n}$?

It seems intuitive, but the empty set seems to be a counterexample of this claim. If the theorem is false, is there any way to salvage it?

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If $\varnothing\ne X\subseteq\Bbb R^n$, there are $2^\omega=|\Bbb R|$ continuous real-valued functions on $X$. Clearly there are at least that many, since for every $r\in\Bbb R$ the constant function $f:X\to\Bbb R:x\mapsto r$ is continuous. Thus, we need only show that there are at most $2^\omega$ continuous functions on $X$.

$\Bbb R^n$ has a countable base for its topology, so it’s hereditarily separable, and therefore $X$ has a countable dense subset $D$. If $f,g:X\to\Bbb R$ are continuous, and $f\upharpoonright D=g\upharpoonright D$, then $f=g$. (This is a standard result, and you might like to try to prove it if you’ve not seen it before: the proof isn’t hard.) Thus, the number of continuous real-valued functions on $X$ is at most the number of real-valued functions on $D$. Since $D$ is countable, this is $|\Bbb R|^{|D|}\le\left(2^\omega\right)^\omega=2^\omega$: there are at most $2^\omega$ continuous real-valued functions on $X$.

Putting the pieces together, we see that there are precisely $2^\omega$ continuous real-valued functions on $X$.

Of course if you don’t specify the codomain of the functions, there are many more than $2^\omega$. As long as the codomain has cardinality $2^\omega=|\Bbb R|$, however, the result holds.

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    If the codomain has cardinality less than $\mid \mathbb{R} \mid$, wouldn't the cardinality of continuous functions on X be less than that of the continuum? For example, if X is a finite subset of $\mathbb{R}$ with the subspace topology, aren't there finitely many continuous functions from X to X?2012-10-17
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    @Erdong: I said nothing about the case in which the codomain has cardinality less than $2^\omega$.2012-10-17
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    Does the codomain need to be Hausdorff for us to conclude that $f = g$ from $f |_D = g |_D$?2012-10-17
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    @ZhenLin: Yes, it does.2012-10-17
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    My question was a bit vague. I am primarily concerned with continuous functions with X as both the domain and codomain. I understand that finite and countable X are counterexamples, but is there a way for the theorem to be true with a looser criterion than the codomain having cardinality of the continuum?2012-10-17
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    If $X$ is Hausdorff, and $\kappa=d(X)$, the minimum cardinality of a dense subset of $X$, then certainly $|X|\le|C(X,X)|\le|X|^\kappa$. If $X$ is discrete, $|C(X,X)|=|X|^{|X|}$, which is $2^{|X|}$ if $X$ is infinite. I’ve not given it a lot of thought, but I suspect that for anything much beyond that you need to know more about $X$.2012-10-17
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    @Brian M. Scott ,Can you prove that an uncountable connected subspace of $R^n$ has cardinal $2^{\omega}$? If so, then I can show that $|C(X,X)|=2^{\omega}$ for any uncountable $X\subset R^n$ such that $|X|<2^{\omega}$.2016-02-02
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    @user254665: Suppose that $C\subseteq\Bbb R^n$ is connected with more than one point. Then each of its $n$ coordinate projections is connected, so each is an interval, possibly degenerate (i.e., a singleton). They can’t all be singletons, however, so at least one is a non-degenerate interval and therefore has cardinality $2^\omega$. Clearly this forces $|C|=2^\omega$.2016-02-02
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    @Brian M Scott. Thanks. I should have thought of that.But I withdraw my claim. I found that I had a hidden assumption that is wrong.2016-02-02
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    @user254665: I know that feeling all too well!2016-02-02