5
$\begingroup$

There is a concept that I have been thinking about quite a lot lately as I am currently self-studying point-set topology:

Say we have a bijective map from one interval, $[a,b]$, to another interval, $[c,d]$, both of which are in $\mathbb{R}$. Also set $c$ and $d$ so that $[c,d] \subseteq [a,b]$.

How can it be that function maps to a subset which is a proper subset of the map's preimage bijectively? I.e. How can the map be both one-to-one and onto when the image should contain "less" elements than the domain?

One example would be $f(x) := \frac{x}{1+x}: [0,10] \to [0, \frac{10}{11}]$

I'm hoping someone can show me why this isn't such a strange concept? Is there a theorem or result that explains this or provides some intuition?

$\endgroup$
  • 2
    $\begingroup$ The existence of a bijection proves that in fact, $[c,d]$ and $[a,b]$ have the same cardinality. $\endgroup$ – Sahiba Arora Jul 14 at 13:31
  • $\begingroup$ @SahibaArora That is exactly what confuses me. The $[c,d]$ is a subset of $[a,b]$ and yet they have the same cardinality. I suppose this is the limit of intuition regarding uncountably infinite sets. $\endgroup$ – samvoit4 Jul 14 at 13:36
  • 1
    $\begingroup$ You should probably spend some time studying Cantor's theory of infinite sets, sometimes found in early parts of analysis books. $\endgroup$ – Justin Young Jul 14 at 14:06
  • 1
    $\begingroup$ I'm mostly just repeating other comments but the main idea is that you've see how certain intuitions for finite sets break down for infinite sets. If I partition a finite set into two nonempty pieces, then our intuition that the pieces are "smaller" is justified. But this breaks down for infinite sets, such as partitioning the set of integers into even and odd. I think once one realizes that they are trying to apply intuition from the finite world, it helps dispel the mystery and things don't seem as strange anymore. $\endgroup$ – halrankard Jul 14 at 14:29
  • 2
    $\begingroup$ Something that might cloud the issue in your particular example is that there is a way to view an interval like $[0,1]$ as "smaller" than an interval like $[0,2]$ via Lesbegue measure. But this is a very different way to "measure" sets compared to cardinality, which is what bijections detect. $\endgroup$ – halrankard Jul 14 at 14:33
7
$\begingroup$

The same applies to any infinite set. In fact "infinite set" can can be defined as a set that contains a proper subset with the same cardinality. More precisely these are known as Dedekind-infinite sets.

For example take naturals $\mathbb{N}=\{0,1,2,3,\ldots\}$ and its proper subset $\mathbb{N}_+=\{1,2,3,\ldots\}$ and note that there's a simple bijection between them $x\mapsto x+1$. You can even remove infinitely many elements from $\mathbb{N}$ and still end up with the same cardinality, e.g. for $2\mathbb{N}=\{0,2,4,6,\ldots\}$ we have a bijection $x\mapsto 2x$ even though there are infinitely many elements in $\mathbb{N}\backslash 2\mathbb{N}$.

And so "being a subset" and "being equinumerous" are loosely related concepts. At most we know that $|A|\leq |B|$ when $A\subseteq B$. But $A\subsetneq B$ doesn't imply $|A|<|B|$, unless $B$ is finite.

I'm hoping someone can show me why this isn't such a strange concept?

The idea may be strange to you. Infinities are weird. But most people simply accept that and move on. There's not really anything more to do about it. After some time you get used to it and it becomes a simple fact of mathematical reality.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Infinites are not a very intuitive concept. In a sense, you can't treat infinity as a value and compare it to other values. You have to talk about properties of the infinity such as whether or not you can "count" everything in that infinity starting with a first element and systematically picking a second element and third element and so on such that you are able to reach all elements. In this case, neither are countable, but you can show that they are both at the same "stage" of infinity. This can be done using a bijection which shows that for every element of the first set you can map it to a distinct element of the second set and vice versa.

If you would like me to clarify, please ask :)

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

Consider as an example there exists a bijection between $\mathbb{Z_+}$ and $\mathbb{Z}$. While these are subsets of $\mathbb{R}$, the concept is the same.

Let $f: \mathbb{Z} \to \mathbb{Z_+}$ defined by

$f(i) = \{1$ if $i = 0$; $2i$ if $i$ is positive; $1-2i$ if $i$ is negative

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How does this answer OP's question? Clearly OP wants to understand the reason why such bijection exists between sets of seemingly different cardinality. $\endgroup$ – Sahiba Arora Jul 14 at 17:05
  • $\begingroup$ @ Sahiba Arora "I'm hoping someone can show me why this isn't such a strange concept? Is there a theorem or result that explains this or provides some intuition?" $\endgroup$ – RJM Jul 14 at 17:25
  • $\begingroup$ @Sahiba Arora Doesn't seem it was so clear to you, when it was first posted: "The existence of a bijection proves that in fact, [c,d] and [a,b] have the same cardinality." – Sahiba Arora 3 hours ago $\endgroup$ – RJM Jul 14 at 17:29
  • $\begingroup$ Hence, I did not post an answer and left a comment. Your example illustrates that this is true with other infinite sets too and not just intervals. However, it does not give "theorem or result that explains this or provides some intuition". $\endgroup$ – Sahiba Arora Jul 14 at 19:06
  • $\begingroup$ @Sahiba Arora Well, I am glad you realized you did not understand well enough to provide an answer. Does the accepted answer show this for just an interval? In your interpretation, anyway. $\endgroup$ – RJM Jul 14 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.