# I Bijection or only injection?

Gold Member
Is the following (or, the following after any minor errors are corrected) a bijection from the unit square S=[0,1]X[0,1] to the line L=[0,1], or only an injection? If only an injection, are the excluded points in L countable?

[1] Let L be identified with the set of real numbers r, 0 ≤ r ≤ 1, whereby r is in unique decimal form 0.r1r2r3.... , whereby any representation as an infinite sequence 0.s1s2......snsn+100000....., where sn≠0 & n≥ 1, is excluded, as it is identified with 0.s1s2......(sn-1)99999..... (0 remains 0.000....)

[2] Let each point in S be identified with the ordered pair (a,b), with a, b∈L ,
a = 0.a1a2a3.... , and
b = 0.b1b2b3.... ,

[3] Then the function is (a,b) to c, with c =0.a1b1a2b2a3b3.... ,
that is, if c= 0.c1c2c3.... then for n≥1 , n, c2n-1=an & c2n=bn.
(or, to put another way, if a = ∑i=1ai×10-i & b = ∑i=1bi×10-i, then c = ∑i=1(ai×10-2i+1 + bi×10-2i)

(Corrections in the details would be welcome.)

#### fresh_42

Mentor
2018 Award
It is a bit cumbersome to read without LaTeX tags, and probably also in general. Here is the Wikipedia page on Hilbert curves: https://en.wikipedia.org/wiki/Hilbert_curve. They are surjective but not injective.

Why are you interested in your specific construction? I don't like decimal representations in this contexts very much. They tend to hide errors and are in my opinion far too specific to represent real numbers. How does your construction look like without the reference to a discrete representation of a continuous object?

Gold Member
Thanks, fresh_42. I am familiar with space-filling curves, but I saw this construction in a couple of posts (but not in a scientific journal, hence I do not provide a reference) which offered it as a proof that the cardinality of the unit interval and the unit square were equal, and on the face of it, it seems to work, but its simplicity raises my suspicions. Hence I am checking on this forum.

Perhaps my use of notation was a bad idea, hence let me describe the function in words:
(a) for a point in the unit square (a,b), take the decimal expressions (but making a caveat for the cases such as 0.0999... = 0.1000....)
(b) construct a point c on the unit line for which the odd digits are from a, and the even digits are from b (with the same caveat).
(c) the function is f((a,b))=c for all (a,b) in the unit square.

#### Mark44

Mentor
I saw this construction in a couple of posts (but not in a scientific journal, hence I do not provide a reference) which offered it as a proof that the cardinality of the unit interval and the unit square were equal, and on the face of it, it seems to work, but its simplicity raises my suspicions. Hence I am checking on this forum.
This is the standard example for showing a bijection from the unit square to the unit line. However, the function that describes this mapping is not continuous.

Gold Member
Thanks very much, Mark44. That answers my question. To show equal cardinality, the bijection need not be continuous, so I am fine with its absence.

#### WWGD

Gold Member
Thanks very much, Mark44. That answers my question. To show equal cardinality, the bijection need not be continuous, so I am fine with its absence.
It would create a contradiction:Continuous bijection between compact and Hausdorff is a homeomorphism. One of the nice obscure results from pointset Topology.

#### jbriggs444

are the excluded points in L countable?

[1] Let L be identified with the set of real numbers r, 0 ≤ r ≤ 1, whereby r is in unique decimal form 0.r1r2r3.... , whereby any representation as an infinite sequence 0.s1s2......snsn+100000....., where sn≠0 & n≥ 1, is excluded, as it is identified with 0.s1s2......(sn-1)99999..... (0 remains 0.000....)
The elements of L all have representations as terminating decimal fractions. As such, they are all rational numbers. The rationals are, of course, countable. So any subset such as L is countable as well.

With a little hand-waving, this means that the set of broken pairings in the described mapping is at most countable. This in turn opens the way for a "slide-everything-down-by-one" style fixup to repair each occurrence of a missing or duplicate pairing.

Dotting the i's and crossing the t's on such a fix-up requires tedious and careful work, but is not very difficult or interesting.

Gold Member
Thanks, jbriggs444. That completes the answer; your answer is as much "dotting the i's and crossing the t's" as is necessary. The notational details are , for my purposes, unnecessary.

WWGD. Thanks for that detail (which is why I wasn't looking for continuity).

"Bijection or only injection?"

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