Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is convex in $X$, then is $Y$ an interval or ray in $X$? The question is from Munkres and this has been answered a lot of time. However, the problems that I am facing are:

*

*How does a ray look like in $\mathbb{R}^3$ or $\mathbb{R}^2$ ?


*What is an interval in an arbitrary topological space $X$?


*How do I think and find out an example that does not satisfy the question ?(I don't want an answer to this question but a hint so that I can find my own answer.)
An example: Consider the dictionary topology in $\mathbb{R}^2$ and consider the convex set $S=\{x^2+y^2<1\}$.We see that this set $S$ is convex. I am not sure whether it is an interval or ray? What is it?
 A: I think that your confusion may arise from the notion of "convex" --- things aren't just 'convex', they're convex with respect to some notion of "interval" (often a vector-space structure) or with respect to some order.
Anyhow, let me answer your question about the lexicographic order on the plane, where $(a, b) < (p, q)$ if $b < q$ or if $b = q$ and $a < p$, which I write out just to fix the order. (Your text may use the opposite convention, but this is the one I am using. If your text uses the other, just swap the first an second coordinate in every interval below.)
The "increasing ray" from a point $(a, b)$ consists of all points $(x, b)$ where $x \ge a$, together with all points $(x, y)$ where $y > b$. In other words, it's the open half-plane above level $b$,together with the half-line to the right of $(a, b)$.
The intersection of two such rays constitutes an interval.
Now let's look at the unit disk, $D$, which is "convex" as a subset of the 2-dimensional vector-space $\Bbb R^2$, and see whether it's convex with respect to the lexicographic order.
Well, the points $(0,0)$ and $(0, .5)$ are both in $D$, so if $D$ is convex, it must contain every point $(x, y)$ between these two. But let's look at the point $P = (100, 0)$. This point is greater than $(0,0)$ in the lexicographic order, but less than $(0, 0.5)$ in that order, so if $D$ were convex, then the point $P$ would have to be in there. But because $100^2 + 0^2 \ge 1$, $P$ is not in $D$.
So your assumption that $D$ is convex (in the order) is mistaken.
