Is there any function that maps all the points on $[0,1]$ to the entire $\mathbb{R}^2$ plane?
Asked
Active
Viewed 76 times
0
-
Sure, both are uncountable and have cardinality $\aleph_1$ – Maximal_inequality Jun 10 '20 at 16:45
-
I see you edited the question but the answer remains the same. – Maximal_inequality Jun 10 '20 at 16:47
-
Agreed. But is there no way to define a function that actually shows this mapping more clearly? – Dhanush Giriyan Jun 10 '20 at 16:47
-
4@Maximal_inequality The cardinality is $2^{\aleph_0}$. It would require the continuum hypothesis to identify this with $\aleph_1$. – Matt Samuel Jun 10 '20 at 16:49
-
You can construct an explicit bijection. I don't remember the details, but it came up recently. Obviously it won't be continuous. – Matt Samuel Jun 10 '20 at 16:50
-
Oh yes, I agree, I sort of missed that. Anyway my point was all of these are in bijective correspondence with $\mathbb {R}$ – Maximal_inequality Jun 10 '20 at 16:52
-
Yes exactly. But what would be the approach to constructing such an explicit bijection? – Dhanush Giriyan Jun 10 '20 at 16:58
-
1The second part of this answer outlines a bijection between $[0,1)^2$ and $[0,1)$. There are bijections between $[0,1)$ and both $[0,1]$ and $(0,1)$; see this question, for instance. From these pieces it’s not too hard to build the bijection that you want. – Brian M. Scott Jun 10 '20 at 17:25