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Given $E \subset \mathbb{R}$, show that there exists $x$ so that both $E \cap (x, \infty)$ and $E \cap (-\infty,x)$ are uncountable.

A very simple statement that seems to be eluding me..

Here are my thoughts so far:

Let $$\mathbb{R} = \bigcup_{n \in \mathbb{N}} (-n,n).$$
Then there exists $N \in \mathbb{N}$ so that $E \cap (-N,N)$ is uncountable. Now, as an open interval, $$(-N,N) = \bigcup_{j \in \mathbb{N}} I_j$$ where $I_j \cap I_k = \emptyset$ for $j \neq k$.

Suppose no such $x$ exists.

Notice that one and only one of the sets $E \cap I_j$ is uncountable then, for if not, we would be able to find such an $x$. Call this set $I_{m_0}$.

As an open interval, $$I_{m_0} = \bigcup_{j \in \mathbb{N}} I_{j_0}.$$ Again, only one such $E \cap I_{j_0}$ can be uncountable call it $I_{m_1}$. Repeating this process indefinitely, we may write

$$E \cap [-N,N] = (\bigcup_{j \in \mathbb{N}} E \cap I_j \setminus E \cap I_{m_0}) \cup (\bigcup_{j \in \mathbb{N}} E \cap I_{j_0} \setminus E \cap I_{m_1}) \cup \ldots \cup (\bigcup_{j \in \mathbb{N}} E \cap I_{j_{k-1}} \setminus E \cap I_{m_k}) \cup \ldots $$

But notice that this is a countable union of countable sets. Ergo, a contradiction, so such an $x$ must exist.

Is this proof correct? Something seems awry.

I'd like to know
i) if this proof is valid.
ii) A better proof. Surely one exists.

Anthony Peter
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    The big problem is that you can’t write $(-N,N)$ as the union of countably infinitely many pairwise disjoint non-empty open intervals: such a union is never an interval. This answer to an earlier question gives a proof of the desired result (and in fact of a much stronger one). – Brian M. Scott Sep 14 '15 at 21:56
  • I’ve sketched one in my answer. – Brian M. Scott Sep 14 '15 at 22:11

1 Answers1

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SKETCH: Let $$L=\{x\in\Bbb R:(\leftarrow,x)\cap E\text{ is countable}\}$$ and $$R=\{x\in\Bbb R:(x,\to)\cap E\text{ is countable}\}\;.$$

We want to show that $L\cup R\subsetneqq\Bbb R$. Suppose that $L\cup R=\Bbb R$.

  • Use the fact that $\Bbb R=\bigcup_{n\in\Bbb N}(\leftarrow,n)$ to show that $R\ne\varnothing$.
  • Use a similar idea to show that $L\ne\varnothing$.
  • Show that if $x\in L$ and $y\in R$, then $x<y$. Conclude that $\sup L$ and $\inf R$ exist.
  • Use an idea like that of the first point to show that $\sup L\in L$.
  • Conclude similarly that $\inf R\in R$.
  • Use the observation that $L\cap R=\varnothing$ to derive a contradiction.
Brian M. Scott
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