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 $a<b$ 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 $a<b$, $a,b,\in \mathbb{Q}$ there would be irrationals so $(a,b) \notin Y$.

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

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Sep 18 at 16:38
  • $\begingroup$ @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? $\endgroup$ Sep 18 at 17:11

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