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I made an observation that for two finite sets $A$, $B$ that most $R \subseteq A \times B$ where $R$ is a function also 'appear to be' non-linear. It got me wondering if this is true in the highly general case of functions from $\mathbb{R}^n$ to $\mathbb{R}^m$.

Let $\Omega$ be the set of a linear functions $f: \mathbb{R}^n \mapsto \mathbb{R}^m$ and $\Xi$ be the set of all non-linear maps $g: \mathbb{R}^n \mapsto \mathbb{R}^m$.

Is there some measure $\mu : Q \mapsto \mathbb{R}_{\geq 0}$ (or other precise way of quantifying the "size" of sets) that shows whether $\mu(\Omega) \leq \mu (\Xi)$?

Arturo Magidin
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Galen
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    The set of linear functions from $\mathbb R^n$ into $\mathbb R^m$ is a finite-dimensional vector space, whereas the set of all functions has infinite dimension. – Taladris Feb 11 '23 at 03:21
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    @EDX The cardinality of $\mathbb{R}$ is $\mathfrak{c}=2^{\aleph_0}$. Whether this equals $\aleph_1$ is the Continuum Hypothesis. And the cardinality of the set of continuous functions from $\mathbb{R}$ to itself is the same as the cardinality of $\mathbb{R}$. – Arturo Magidin Feb 11 '23 at 03:42
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    Yes thanks for your precision. – EDX Feb 11 '23 at 04:11
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    @EDX in re your deleted answer, you asserted that the cardinality of $\mathbb{R}$ was $\aleph_1$, and that the cardinality of $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ (which is the set of continuous functions from $\mathbb{R}$ to itself) was $2^{\aleph_1}$. So you definitely claimed the set of continuous functions had larger cardinality than the reals. It wasn't a problem of heuristics vs "theoretical proof", the problem is you said a lot of things that were just plain false, not merely imprecise. – Arturo Magidin Feb 11 '23 at 04:17
  • Yes of course that was false. Was a confusion about. – EDX Feb 13 '23 at 09:41
  • Currently there are (+3/-2) votes on this post. I'm trying to understand why there are downvotes, but that is difficult without feedback. I don't mind the downvotes per se, but I would also appreciate feedback on how to improve the post. – Galen Mar 05 '23 at 16:10

2 Answers2

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The set of linear functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ is in correspondence with the set of $m\times n$ matrices with real coefficients. This is a basic result of Linear Algebra. This set has the same cardinality as the reals, namely $\mathfrak{c}=2^{\aleph_0}$.

The set of all functions from $\mathbb{R}$ to itself has cardinality $|\mathbb{R}|^{|\mathbb{R}|} = 2^{\mathfrak{c}}$, which by Cantor's Theorem is strictly larger than $\mathfrak{c}=|\mathbb{R}|$. The same is true for the set of all functions from $\mathbb{R}^n$ to $\mathbb{R}^m$.

Arturo Magidin
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  • This shows that there are functions that are not linear, but it's not obvious to me that it follows that there are more non-linear functions than linear functions. (Although I imagine it would be obvious if I knew how to do arithmetic with cardinals.) – d_b Feb 11 '23 at 16:53
  • @d_b The rest of the argument is alluded to via mentioning Cantor's theorem. There is a short proof on Wikipedia: https://en.wikipedia.org/wiki/Cantor%27s_theorem#Proof – Galen Feb 11 '23 at 16:58
  • @d_b Also on Proof Wiki – Galen Feb 11 '23 at 16:59
  • @d_b And another important aspect, perhaps the one you were looking for: Here we are considering two sets to be of equal "size" (cardinality) if there exists a bijection between them. This definition coincides with how we consider it in finite cases, but is still defined in infinite cases. A bijection requires both injectivity and surjectivity. Cantor's theorem is telling us we do not have the latter. – Galen Feb 11 '23 at 17:06
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    @d_b That there are functions that are not linear is trivial: a constant function that is not the zero function is not linear. The argument above is irrelevant to this. The fact that the cardinal is bigger shows that it is impossible to establish a one-to-one correspondence. Cf. there are strictly more reals than natural numbers in the same sense. – Arturo Magidin Feb 11 '23 at 17:29
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    It might be worth adding the cardinality of the set of continuous functions , which I believe is $\mathfrak{c}$ since you can describe each one by its values on a dense grid of rational points. So most functions are discontinuous – Henry Feb 12 '23 at 02:35
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The other answers provide a nice cardinality-centric answer to the question. As an alternative, I don't believe it's particularly hard to show that the set of linear functions is a closed meager subset of the topological space of continuous functions $C(\mathbb{R}^n,\mathbb{R}^m)$ with the compact-open topology.