4

I am given the following problem:

Let $f$ be a function from $[0, 1]$ to $\mathbb{N}$. Prove there exists $x, y\in [0, 1]$ such that $x\neq y$ and $f(x) = f(y)$.

Clearly, this is true because $|[0, 1]| = |\mathbb{R}| > |\mathbb{N}|$, and there cannot be an injective function from an uncountable set to a countable set. But... how should I write my solution? They are asking for a proof and I do not know how to formally prove the statement above.

Thank you!

Martin Sleziak
  • 56,060
ketsi
  • 3,904
  • 2
    What are you allowed to use? – Mike Mar 06 '22 at 20:41
  • @Mike The thing is I’m not exactly sure, so I am trying to make it as rigorous as possible. Is it okay to simply state that there is no injective function from A to B iff |A| $>$ |B| – ketsi Mar 07 '22 at 00:39
  • 1
    How have you defined the cardinality of a set? Have you already shown that $[0, 1]$ is uncountable? –  Mar 07 '22 at 07:51
  • @LewisMacRae Thank you for your help, I think I don't even know what I'm asking for at this point haha, I will delete my question in a minute – ketsi Mar 07 '22 at 07:53
  • The question is fine, in my opinion. I reckon you've already answered it. If there was an injection $[0, 1] \to \mathbb{N}$, then $[0, 1]$ would be at most countably infinite. But this contradicts the uncountability of $[0, 1]$. –  Mar 07 '22 at 07:59
  • 1
    @LewisMacRae you say, "If there was an injection $[0,1]\to\mathbb{N}$ then $[0,1]$ would be at most countably infinite." Whilst this seems obvious, it still requires justification. How do you justify it formally? – Adam Rubinson Mar 07 '22 at 14:10
  • 2
    What about this argument: $g:\mathbb{N}\to [0,1],\ g(n) = \frac{1}{n}$ is an injection. If an injective function $f:[0,1]\to\mathbb{N}$ existed, then by Schröder–Bernstein theorem, There would exist a bijection from $[0,1]\to \mathbb{N},$ contradicting Cantor's Diagonal Argument. https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem – Adam Rubinson Mar 07 '22 at 14:16
  • 1
    @AdamRubinson Hi Adam. If you define cardinality via functions, my statement is essentially by definition (see wiki). I think your argument is excellent - I love the Schröder–Bernstein theorem! –  Mar 07 '22 at 14:38
  • @LewisMacRae Actually I wrote up an answer that does not use CSB! – Vivaan Daga Mar 14 '22 at 06:34
  • @AdamRubinson Notifying you of the above. – Vivaan Daga Mar 14 '22 at 06:36
  • This question seems a bit related: Prove that there is no injection from $\Bbb{R}$ to $\Bbb{N}$. – Martin Sleziak Mar 14 '22 at 07:27
  • Let $A\subseteq\mathbb N$ be an image of $f$. Then $f^{-1}: A\to\mathbb[0,1]$ is a surjection. Repeat a proof by Cantor to show such $f^{-1}$ does not exist. – CiaPan Mar 14 '22 at 08:47

1 Answers1

1

Consider $2^\omega$, the set of all binary sequences, one can inject $2^{\omega}\to [0,1]$, by mapping each binary sequence $x_{n}$, to $\sum_{n=0}^{\infty} \dfrac{x_{n}}{10^{n+1}}$ which belongs to $[0,1]$, it’s easy to check that this mapping is an injection.

So if there was an injection from $[0,1]$ into $\mathbb{N}$, there would be a surjection from $\mathbb{N}$ onto $2^\omega$, a simple diagonalization argument shows that is impossible for there to be such a surjection, therefore we have a contradiction. $\square$

Vivaan Daga
  • 6,072
  • Neat, but why bother with binary sequences? If there was an injection $[0, 1] \to \mathbb{N}$, then there would be a surjection $\mathbb{N} \to [0, 1]$ -- but this contradicts Cantor's diagonal argument. –  Mar 14 '22 at 09:27
  • @LewisMacRae What do you mean by it "contradicts Cantor’s diagonal argument"? The usual cantor argument is made for binary sequences and I showed that the binary sequences inject into $[0,1]$. – Vivaan Daga Mar 14 '22 at 12:05
  • Please ignore my last comment! The diagonal argument I remember learning years ago used the decimal representations of the real numbers in $[0, 1]$, but I have just read that Cantor originally considered binary sequences. TIL. –  Mar 14 '22 at 12:11
  • @LewisMacRae Note that if you use decimal expansion-then the argument is somewhat sticky-you have to prove that every number has a decimal expansion-and then you have to be careful about the number you get from the diagonalization since some numbers have two expansions. (Some texts do this wrongly) – Vivaan Daga Mar 14 '22 at 12:37
  • My answer avoids both issues. – Vivaan Daga Mar 14 '22 at 12:37
  • See also https://math.stackexchange.com/questions/2094116/how-do-we-know-that-cantors-diagonalization-isnt-creating-a-different-decimal/2094120#2094120, and specifically Noah Shweber’s comment. – Vivaan Daga Mar 14 '22 at 12:38