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Let $A$ be an uncountable subset of $\mathbb{R}$. Prove that there exists an integer $m$ such that the interval $[m,m+1]$ contains uncountably many elements of $A$.

I can prove this by defining $A_m = A \cap [m,m+1]$ for every $m \in \mathbb{Z}$. Since $A = \bigcup_{m \in \mathbb{Z}} A_m$, there must be at least one $A_i$ that is uncountable -- otherwise $A$ would be countable.

My question is what are the candidates for $A$? Are the irrationals (or subsets of the irrationals) and real intervals the only uncountable subsets of $\mathbb{R}$?

wildcat
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    $A$ can be many other things. $\Bbb{R}$ is uncountable, but is not a subset of irrationals. Perhaps what you meant to ask is "must every uncountable subset of $\Bbb{R}$ contain an irrational number?" and the answer to this is yes (in fact it must contain uncountably many irrationals, but not necessarily all of them). This is simply because $\Bbb{Q}$ is countable. – peek-a-boo Dec 17 '21 at 00:07
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    What happens when you remove an at-most countable set from $\mathbb R$? – John Douma Dec 17 '21 at 00:08
  • $\mathbb{R} \setminus \mathbb{Q} = I$. So $A$ must contain uncountably many irrational numbers, because otherwise $A$ would be a subset of $\mathbb{Q}$ and thus countable. Ok, that makes sense. Thank you! – wildcat Dec 17 '21 at 00:15
  • The most popular uncountable subsets of $\mathbb R$ would be the perfect sets, i.e., nonempty closed sets that contain no isolated points. Leading the polls among those is the Cantor ternary set. It is only calculus students that would vote for the irrationals. – B. S. Thomson Dec 17 '21 at 00:18
  • In particular, every uncountable real subset must contain a transcendental number. It's not hard to show that the set of algebraic reals is countable. – While I Am Dec 17 '21 at 00:30
  • $A$ can still contain rational numbers though. – user_194421 Dec 17 '21 at 01:52

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Maybe this is close to your thinking:

Take a set $A\subset \mathbb R$ that is uncountable and take any subset $B\subset A$ that is countable. Then the set $A\setminus B$ is uncountable. Do all examples of uncountable sets look much like this? Is this the only method of finding such examples?

So for example, we know an interval $[a,b]$ is uncountable. So a simple example of a relatively interesting uncountable set is $[a,b]\setminus S$ where $S$ is some set that I can reasonably show is countable (e.g., all rationals, all algebraic numbers, the range of some sequence, etc.)

But there are much more interesting examples of uncountable sets. Every perfect subset of $\mathbb R$ (i.e., every nonempty closed set without isolated points) is uncountable and in fact has cardinality $\mathfrak c$ of the continuum. All of the Cantor-like sets (including the orginal Cantor ternary set) are perfect and so uncountable.

For this see: Proof that a perfect set is uncountable

If you have studied Lebesgue measure on the real line then you have an abundance of interesting uncountable sets. If $E\subset \mathbb R$ is a measurable set of positive Lebesgue measure then it is trivial to show that $E$ is uncountable. This cheats a bit since every such set $E$ contains a perfect set so we are back to saying it is both uncountable and has cardinality $\mathfrak c$.

Of course there are plenty of other ways an uncountable set might arise naturally. For example let $f:[a,b]\to \mathbb R$ be a continuous function with the property that every level $$f^{-1}(L)= \{x\in [a,b]: f(x)=L\}$$ is countable (i.e., finite or countably infinite). Then there is an uncountable set of points where the derivative $f'(x)$ must exist.

It follows, too, that if $f$ has a derivative at only a countable number of points then there are uncountably many values of $L$ so that the levels $f^{-1}(L)$ are themselves uncountable. (By the way, this leads us back to perfect sets: levels of a continuous function are always closed and most continuous nowhere differentiable functions must have most levels perfect.)

In short you will, before long, see more instances of uncountable sets than you might care to confront.

B. S. Thomson
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There are lots of uncountable subsets of $\Bbb R$. There are $2^{\mathfrak c}$ of them because there are $2^{\mathfrak c}$ subsets of $\Bbb R$ and only $\mathfrak c$ of them are countable. You and I can only imagine a vanishingly small fraction of them. We have to prove things about them without imagining the variety of them.

Ross Millikan
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