Thinking of Hilbert's Hotel (https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel). Let's assume we take an infinite set $S$ of and we assign a unique name to every element in $S$. From that we get a set of all unique names in $S$ called $N$. If I would remove a single element from the set $S$, can I rename all the remaining elements in $S$ so that the set $N$ remains 'unchanged'?

Edit: As pointed out in the answers below the original formulation that $S$ is a bounded subset of ${\rm I\!R}^1$ is obsolete.

  • $\begingroup$ For an arbitrary bounded subset? What if it's finite? $\endgroup$ – Kevin Long Sep 22 '17 at 19:50
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    $\begingroup$ The short answer is: "If and only if $S$ is infinite". The longer answer (since you appear to be fond of hand-wavy things rather than the real deal) is: "It depends on the meaning of the word can". $\endgroup$ – user228113 Sep 22 '17 at 19:53

If there is a one-to-one function $f:\mathbb N\to S$, then you can do this. Let $I=f(\mathbb N)$.

Then if you remove $x_0\in I$, then $x_0=f(n)$ for some unique $n$. Then define the new names $N_2$ on $X\setminus\{x_0\}$ as:

$$N_2(x)=\begin{cases}N(f(m))&x=f(m+1)\text{ with }m\geq n\\ N(x)&\text{otherwise}\end{cases}$$

If $x_0\notin I$ then we define:

$$N_2(x)=\begin{cases}N(x_0)&x=f(0)\\ N(f(m))&x=f(m+1)\\ N(x)&\text{otherwise} \end{cases}$$

Now, if we know $S$ is infinite, we can prove there is an $f:\mathbb N\to S$ by defining $f$ inductively.

There is really no reason to add the condition that $S$ be bounded, or even that $S$ is a set of real numbers. The only question is whether $S$ is infinite or not.

  • $\begingroup$ Ok, for clarification. Suppose we have an infinite set $S$ such that there is a one-to-one function $f:{\rm I\!R}\rightarrow S$. Can I still play the same game? $\endgroup$ – piiipmatz Sep 23 '17 at 6:34
  • $\begingroup$ Sure, because then you have a one-to-one function $f:\mathbb N\to S$. $\endgroup$ – Thomas Andrews Sep 23 '17 at 13:57

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