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I understand the proof based on ordering $[0,1]$ into a set $A$ of distinct points that include $0$ and $1$ and then showing the one-to-one equivalence to $(0,1)$, but what I can't get my head around is that from a non-mathematical view, it is clear that $[0,1]$ contains precisely $2$ more elements than $(0,1)$?

I expect it has something to do with cardinality, and I get the basic concepts of equivalence and cardinality, but still trying to wrap my head around this question. I’m comfortable with the equivalence of $[0,1]$ to $[0,2]$ for example.

This question is very similar but slightly different to this one: Are all infinities equal?

I am specifically looking at the equivalence between open and closed sets. For example, if I map every element $x\in (0,1)$ to every element $y \in[0,1]$ such that $x = y$ then I will be left with two extra elements in $[0,1]$. To me, this breaks one-to-one correspondence in how its typically interpreted, however I can see that it does not actually break the definition if we follow it to the strict letter since the idea of leftover elements does not factor into the definition: https://en.m.wikipedia.org/wiki/Bijection

I’m definitely in the wrong here but I’m trying to understand the intuition.

Ashok Khanna
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  • Why are you comfortable with the latter which by the same logic has infinitely many more points? – Peter Foreman Apr 16 '20 at 23:31
  • For the bijection between $[0,1]$ and $[0,2]$, just consider the multiplication-by-$2$ map. – Geoffrey Trang Apr 16 '20 at 23:33
  • I’m comfortable due to the multiplication by 2 map, but here I’m not 100% comfortable between open vs closed even though I understand the proof. I’ve tried to create another proof to contradict the equivalence - I’m missing something here – Ashok Khanna Apr 16 '20 at 23:41

1 Answers1

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The open set $A=\{x|x\in \mathbb{R},0<x<1\}$ and the closed set $B=\{x|x\in \mathbb{R},0\le x \le 1\}$, are distinct sets. One is strictly a subset of the other, and yes $B-A=\{0,1\}$. This can be true despite having the same cardinality.

It turns out it is impossible to have a continuous bijection from a closed set to an open set, but there are non-continuous ones. We have a finite number of elements we need to add. I'm sure you've seen the "Hilbert hotel" trick of sliding everyone down to fit one more element in. So all we need to do is select two "countably infinite" subsets, and slide them down one.

$$f : [0,1] \rightarrow (0,1)$$ where f(0) = 1/2, f(1) = 1/3,
if $x=2^{-n}$ for some integer $n>0$ then: $f(x) = x/2$,
if $x=3^{-n}$ for some integer $n>0$ then: $f(x) = x/3$,
else $f(x) = x$

While not continuous, this is invertible, therefore providing a bijection between [0,1] and (0,1).

HBrown
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  • So then, $[0,1]$ and $(0,1)$ are not homeomorphic, right? – Geoffrey Trang Apr 17 '20 at 13:45
  • Thanks for this. It points me in the right direction with the lack of continuity of the bijection from closed set to open set. Would you be able to expand on the implications of A - B = {0,1} (a finite set) and the implications for cardinality? I can get my head around C = [0,1], D = [0,2] such that D - C = (1,2] being a set with infinite elements and thus not really countable in the plain english terminology and accordingly the concept of one-to-one correspondence makes more sense to me, but do you have any extra thoughts on the analogous example but where the complement is a finite set – Ashok Khanna Apr 17 '20 at 14:46
  • @4d65 Consider the set of non-negative integers, and the set of integers > 1. The difference is {0,1}, and both sets have the same cardinality. Does that feel more comfortable? Ultimately this is an intuition thing, that we all struggle with slightly differently. Playing with infinities is pretty far from everyday intuition. – HBrown Apr 17 '20 at 15:56
  • @GeoffreyTrang My understanding is that you are correct. I don't personally know how to prove a continuous bijection is impossible, so it would be best to ask that as a separate questions to get a more detailed answer. – HBrown Apr 17 '20 at 16:01
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    @geoffreytrang https://math.stackexchange.com/questions/2862117/prove-that-an-open-interval-0-1-and-a-closed-interval-0-1-are-not-homeomorph – PersonX Apr 17 '20 at 16:27
  • @HBrown Thanks, that helps. I am comfortable with that, and accordingly, while still not 100% there, can appreciate the overall answer more fully. Will mark it as closed. Thanks! Perhaps worth adding that part to the answer for others to see – Ashok Khanna Apr 17 '20 at 17:04