# [Real Analysis] Show that the set $A$ is uncountable. Use this result to show that ${\displaystyle\mathbb {R}}$ is uncountable.

Class exercise:

Let $A$ be the set of all real sequences whose elements are just 1, 2 or 3, i.e.
$A=\{\{x_n\} ? {\displaystyle\mathbb {R}} : x_n ? \{1, 2, 3\}, ?n ? {\displaystyle \mathbb {N}}\}$.
a) Show that $A$ is uncountable.
b) Use this result to show that ${\displaystyle\mathbb {R}}$ is uncountable.

Attached is part of the material used, for reference on the notation. We're probably meant to extend Theorem 1.10.11 on page 45, but I can't quite figure it out.
• Can't you just use Cantor's diagonalization argument? You can prove a & clearly there is a bijection w/ the numbers in [0,1] with digits after decimal 1,2,3 so that is uncountable hence R which includes that is uncountable. I don't think it takes grad level real analysis

• Yeah, I forgot to mention that. I don't think we can use the argument...

• Let me check.

• Yep, wasn't announced to us, unfortunately.

• But, was it explicitly stated you weren't allowed to?

• It's not a very high powered technique, so unless you were explicitly told you weren't allowed to, you should safely be able to use it.

• Just asked the professor. We can in fact use it, sorry for the inconvenience. Would you be able to break it down in the answer? I really want to understand it properly.

• OK sure. I will try to. I felt that Cantor's diagonalization argument was most natural.

• OK I have a decent writeup now which contains the critical details, assuming some relatively simple facts.

• The argument I used in (a) is also what is called Cantor's diagonalization argument by the way, if that was not clear

Answers can be viewed only if
1. The questioner was satisfied and accepted the answer, or
2. The answer was disputed, but the judge evaluated it as 100% correct.

2 Attachments

• If the decimal expiation of number has a pattern, it is a rational number. Let $B=A \ R$ where $R$ is the set of sequences who have a pattern. Then B is uncountable and the given map in part B becomes one to one. Statements like "in which case the proof would look more complicated but still works out due to how ”sparse” these limiting equivalences are)" should not be part of a formal proof. Please revise your solution accordingly and upload it in a separate attachment.

• I suppose this portion could be expressed more precisely. As I mentioned, I don't believe it is relevant to this particular proof since I don't believe we actually need to take away anything from A to obtain a bijection in this case. Hence, why it's not important to consider for this particular proof. There also seems to be imprecision in your description as well which obscures my interpretation of those comment (in regards to Sequences w/ pattern, though rationals are familiar). Please clarify.

• "I don't believe we actually need to take away anything from A to obtain a bijection in this case" should be backed up with a rigorous proof. My comment was just a hint to you and of course needs to be polished a bit, but it actually shows what you meant by sparse in your argument.

• OK. I thought this was a minor detail that could be filled in by the reader & that I highlighted what might be considered the major points of the situation. I suppose I can expand then if necessary.

• Your solutions is very well-written that was the only place where there was some room for more details/explanation.

• I have described this in slightly more depth (it is about particular rational numbers rather than in general actually but subtracting the sequences mapping to rationals also works I think), though there is always some level of imprecision that must be accepted as it is never possible to provide a complete description of an object as we must rely on consistent, finitely generated description of properties of the object & I still wish to provide the reader with some amount to think about.

• Looks great now. Thank you for adding those details.