Counter argument for the characterization of intervals.

Question

My text has an Interval Characterization theorem as follows:

If $S$ is a subset of $\mathbb{R}$ that contains at least two points and has the property $$\text{ if }\, x,y \in S\, \, \, \text{ and } \,\,\, x< y\,, \,\,\, \text{ then } [x,y] \subseteq S \tag{1}$$ then $S$ is an interval.

I say this is false. Consider the subset $S$ in $\mathbb{R}$ defined $S = \{3\} \cup[4,5] \cup \{6\}$. Clearly $4,5 \in S$ and $4<5$ then $[4,5] \subseteq S$.

But $S$ is not an interval. So where am I going wrong here?

Answer 1

You don't understand the statement. It says:

Given any imaginable set $S$ that fulfills two certain conditions, we can say that $S$ is an interval. The conditions are 1. $S\subseteq \mathbb{R}$ and 2. $\forall x,y\in S:x<y\Rightarrow [x,y]\subseteq S$.

Your example isn't an interval, which we both see, but it is a subset of $\mathbb{R}$, so condition 2 must be false. It is, because it does not hold for $x=3$ and $y=4$, but is has to hold for any two element of $S$. Nothing is wrong with the theorem itself, it is still true.

In other terms, your example doesn't fulfill condition 2, so it is no counterexample to the theorem.