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If $X$ is uncountable, then the set of bounded injective functions: $f: X \rightarrow \mathbb{R}$ has empty interior.

Attempt:

In order to prove that the interior is empty, we must show that for any $f$ in the set and for any ball centered on $f$: $B(f;\epsilon)$, one can find a function $g\in B(f;\epsilon)$ such that $g$ is not injective.

Take any $f$ in the set and any $\epsilon>0$. We note that the image of $f$, $f(X)$, is an uncountable subset of $\mathbb{R}$. Therefore, it contains at leats one of its accumulation points, say $f(x_0$), where $x_0\in X$. Hence it follows that $B(f(x_0);\epsilon/2)\cap f(X) \neq \emptyset$. We take $f(y_0)\in B(f(x_0);\epsilon/2)\cap f(X)$ such that $f(x_0)<f(y_0)$, where $y_0\in X.$

Define the following function:

$g:X\rightarrow \mathbb{R}$ by letting: $g(y_0) = f(x_0)$ and $g(x)=f(x)$ elsewhere. Clearly $g$ is not injective and:

$$\mbox{sup}_{x\in X}|f(x)-g(x)| = |f(y_0)-g(y_0)|=|f(y_0)-f(x_0)|<\epsilon,$$

since we chose $f(y_0)\in B(f(x_0);\epsilon/2)$. This shows that $g\in B(f;\epsilon)$, $g$ being not injective.

Is this fine? Any different idea?

user2345678
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  • You need to tell us what topology (or metric) you are using on the space of (bounded?) functions $X \to \Bbb{R}$. – Rob Arthan Sep 09 '17 at 22:45
  • @RobArthan The metric of sup: $d(f,g)=\mbox{sup}_{x\in X}|f(x)-g(x)|$

    Also this is the space of bounded real functions.

     – user2345678 Sep 09 '17 at 22:48
  • You should put that clarification in the question. – Rob Arthan Sep 09 '17 at 22:55
  • I think when you say $B(f(x_0);\epsilon/2)\cap f(X)$ you mean to remove $x_0$, that is, $(B(f(x_0);\epsilon/2)\cap f(X)) \setminus {x_0}$ – Dap Sep 09 '17 at 22:56

1 Answers1

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Your proof is good, though as a comment mentioned you should be clearer what topology (and ambient space) you're using.

Another idea is to define $g(x)=\epsilon\lfloor f(x)/\epsilon\rfloor$ for all $x$. This gives $|f(x)-g(x)|<\epsilon$ for all $x$ as in your argument. But the image of $g$ is finite: it's a subset of $\{-N\epsilon,-(N-1)\epsilon,\dots,N\epsilon\}$ for some bound $N$. So by the pigeonhole principle $g$ cannot be injective.

Dap
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