Do exist an injective function from $\mathbb{R}^5 \rightarrow \mathbb{R}^4$?

I think it is true, but I am unable to find a simple example... Please help me.

I think it should be something like $$ q:\mathbb{R}^5 \rightarrow \mathbb{R}^4, \ q(x,y,z,k,l) = (f(x,y),\ f(y,z),\ f(z,k),\ f(k,l)); $$ where $f:\mathbb{R}^2\rightarrow\mathbb{R}$, an injective function, as I just read here about that: Injective function from $\mathbb{R}^2$ to $\mathbb{R}$?

Am I wrong?

If there exists a such function, please give me an example. Thank you very much!

  • 5
    $\begingroup$ Once you know an injective function $f$ from $\mathbb{R}^2$ to $\mathbb{R}$, you can just use $\Phi((x,y,z,k,l))=(f(x,y), z, k, l)$ $\endgroup$ – Evargalo Jun 19 '17 at 14:13
  • 3
    $\begingroup$ There exists a bijection between $\mathbb R^5$ and $\mathbb R^4$, because these sets have the same cardinality (equal continuum). $\endgroup$ – A.B Jun 19 '17 at 14:17
  • $\begingroup$ Thank you all! You helped me a lot! $\endgroup$ – MM PP Jun 19 '17 at 14:21
  • 2
    $\begingroup$ It should be noted, there is no continuous injection. $\endgroup$ – Thomas Andrews Jun 19 '17 at 14:25

You can use the injection from $\Bbb R^2 \to \Bbb R$ that you read about. If that bijection is $m=f(x,y)$ you can take $(x,y,z,k,l) \to (m,z,k,l)$ The last three components just go along for the ride.

  • $\begingroup$ Thank you! You helped me a lot! $\endgroup$ – MM PP Jun 19 '17 at 14:21

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.