# Question about convexity of ordered set

This is an example from Munkres' Topology.

The definition given states that for an ordered set $$X$$, let us say that a subset $$Y$$ of $$X$$ is convex in $$X$$ if for each pair of points $$a of $$Y$$, the entire interval $$(a,b)$$ of points of $$X$$ lies in $$Y$$.

Now I found a counterexample to the statement that if $$Y$$ is a proper subset of $$X$$ that is convex in $$X$$, it may not follow that $$Y$$ is an interval or a ray in $$X$$.

Here the set $$Y$$ is $$\mathbb{Q}$$ and $$(-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$$ is given as a convex subset which is not an interval in $$\mathbb{Q}$$.

My concern is why is the set $$Y$$ convex in the first place? For every $$a, $$a,b,\in \mathbb{Q}$$ there would be irrationals so $$(a,b) \notin Y$$.

• Note that some people define an interval as a convex subset. So this would be an interval according to that definition. Sep 18 at 16:25

For any two rational points $$a,b \in Y$$ the interval $$(a,b)$$ is in $$Y$$ making it convex. In this example $$X= \mathbb{Q}$$ and so $$(-\sqrt{2},\sqrt{2}) \cap \mathbb{Q}$$ doesn't contain any irrational numbers. The reason it's not an interval is that the endpoints are not in $$\mathbb{Q}$$. It is the union of infinitely many rational intervals though.
• I am confused a little here. The problem asks to show that if $Y$ is a convex subset in $X$ then it may not be an interval. You take $X=\mathbb{Q}$ and $Y=(-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$. But this $Y$ is not convex since $-\sqrt{2}$ and $\sqrt{2}$ does not belong to the rationals. So how does this prove the question? Sep 18 at 16:38
• @nomadicmathematician It's convex because for any two rational numbers $-\sqrt{2} < a \leq b < \sqrt{2}$ the interval $(a,b) \subset Y$. Can you see why? Sep 18 at 17:11