I found the following proof to be constructive:

  1. There is a bijection from $[0,1]$ to $(0,1]$. Have $0\mapsto \frac12, \frac12\mapsto\frac23,\frac23\mapsto\frac34,$ and so on. That takes care of $\left\{0, \frac12, \frac23, \frac34,\ldots\right\}$. For any other $x$, just map $x\mapsto x$.

This is from https://math.stackexchange.com/a/183383/1125430 .

However, it seems to be using the law of excluded middle $x\in \left\{0, \frac12, \frac23, \frac34,\ldots\right\}\vee x\notin \left\{0, \frac12, \frac23, \frac34,\ldots\right\}$ to define the function.

Is the above proof constructive?

  • 1
    $\begingroup$ It is constructive because it defines the function to be considered. Any concerns about the use of the law of excluded middle are irrelevant to whether this is considered a constructive proof or not. If you actually have concerns about the use of excluded middle, then your concern might be whether the proof was valid or not... but the vast majority of people accept working in a logic system where the law of excluded middle is assumed and so would have no qualms about it appearing here. $\endgroup$
    – JMoravitz
    Jun 9 at 12:04
  • $\begingroup$ Admittedly, the "and so on" is informal and should have been written explicitly as $1-\frac{1}{n}\mapsto 1-\frac{1}{n+1}$... so that the continued pattern is unambiguous. $\endgroup$
    – JMoravitz
    Jun 9 at 12:06
  • $\begingroup$ Even a mathematician not accepting the excluding middle would see no problem here. We have two sets such that obviously every real number in the interval $[0,1]$ must be in exactly one of the sets. It is anyway a strange idea to doubt the law of the excluding middle , which is a very important and powerful law , needed for almost every interesting result. $\endgroup$
    – Peter
    Jun 9 at 12:25
  • 3
    $\begingroup$ The proof is not constructive if your intervals are understood to be in $\mathbb{R}$, but it is if they are understood to be in $\mathbb{Q}$ or in some well-understood extension of $\mathbb{Q}$ (I am being deliberately vague here, since mere decidability of equality is not sufficient; you need to be able to tell whether a given $x$ belongs to $\left\{0, \frac12, \frac23, \frac34,\ldots\right\}$, and if yes, to determine the $n$ satisfying $x = \dfrac{n}{n+1}$). $\endgroup$ Jun 9 at 13:24

1 Answer 1


This is not an acceptable proof in constructive mathematics, because it relies essentially on the law of excluded middle.

One can use your argument to construct a function with domain $P \cup [0,1]\setminus P$ and codomain $(0,1]$ where $P = \left\{0,\frac{1}{2},\dots \right\}$, and that in itself is an acceptable construction in constructive mathematics. But unfortunately you cannot show that $P \cup ([0,1]\setminus P) = [0,1]$ without using (a non-constructive instance of) the law of excluded middle.

You never actually write down an argument showing that the function you construct is a bijection -- and if you did, you'd find that showing this requires another use of the law of excluded middle.

In fact, using arguments that are acceptable in all flavours of constructive mathematics, it's not possible to construct a discontinuous function $[0,1] \rightarrow [0,1]$ at all! And the function you define above is clearly discontinuous (NB this is a very hard constraint for your problem: as an exercise, think about compact sets, and whether it's possible to have a continuous bijection between an open and a closed interval in classical mathematics.)


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