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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!

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ It should be noted, there is no continuous injection. $\endgroup$ – Thomas Andrews Jun 19 '17 at 14:25
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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.

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

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