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I want to make sure I am understanding this correctly. If we have a function $f$ which maps $A$ to $B$, and $A$ is a subset or proper subset of $B$ then $f$ is one to one from $A$ to $B$? Is that true?

Or does that only hold in the case of $f(x)=x$?

My textbook is trying to show that the $|(0,1)| = |(0,1]|$. I sort of understand the first part:

1) Because $(0,1) ⊂ (0,1]$, $f(x)=x$ is a one to one function from $(0,1)$ to $(0,1]$.

Then it says:

2) The function $g(x)=x/2$ is clearly one-to-one and maps $(0,1] $to $(0, 1/2] ⊂ (0,1)$.

By Schroder-Bernstein theorem, $|(0,1)| = |(0,1]|$

Yet I don't get how the second part is sufficient as the function $g$ maps $(0,1]$ to a subset of $(0,1)$ but forgets about $[1/2, 1)$.

My instinct tells me that these two have different cardinalities, since one includes "1" while the other does not...

Martin Sleziak
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ohbrobig
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2 Answers2

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You have shown that $f:(0,1)\to(0,1]$ is injective and that $g:(0,1]\to(0,1)$ is injective. By the Schroder-Bernstein theorem, there is a bijection $h:(0,1)\to(0,1]$, and so the two sets have the same cardinality.

If you have any confusion, I would recommend verifying that these two functions are, indeed, injective, and maybe also taking a look at the proof of the theorem to see what its implications are.

Rocket Man
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Although $A$ is a proper subset of $B$, the function from $A$ to $B$ need not to be one-to-one, even if $f$ is onto. For example you can consider $A = (-\pi/2,\pi/2)$, $B=\Bbb{R}$ and $f(x) = \tan x$.

You can examine the proof of Schroder-Bernstein theorem. It 'constructs' a bijection between $A$ and $B$ from two one-to-one functions $f:A\to B$ and $g:B\to A$. So the proof of $|(0,1)| = |(0,1]|$ does not forget about $[1/2,1)$, though it looks like be.

Moreover, removing a finite subset from an infinite set does not affect its cardinality. For example, we can construct a bijection between $\Bbb{N}$ and $\Bbb{N}-\{1\}$.

Hanul Jeon
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  • Ok that clarifies things. Is it because g actually map B to a subset of A? And since this mapping is one to one, the mapping of B to A is also one to one? (If every element in the subset is mapped to by one unique element in B, there cannot be any elements outside of the subset, and in A, which are mapped to by B) – ohbrobig Oct 10 '15 at 03:02
  • @ohbrobig Sorry, I can't understand your question. Let me know what "the mapping of B to A" intended. (Does you mean it as the bijection between $B$ and $A$ given from $f$ and $g$ by Schroder-Bernstein theorem?) – Hanul Jeon Oct 10 '15 at 03:10
  • I am referring to map of g (from B to a subset of A) – ohbrobig Oct 10 '15 at 03:38
  • @ohbrobig Extending codomain does not affect the fact that $g$ is 1-1, so, right. The function is one-to-one. – Hanul Jeon Oct 10 '15 at 04:06