Does $[0,+ \infty)$ counts for a closed interval in just real number system?

This is different from my previous question which asked about the situation in extended real number system.

So, we are discussing the situation when $$X=\mathbb{R}$$. Without the elements $$\{-\infty, +\infty \}$$, we could just define $$[0,+\infty)$$ formally to be $$\{t\in X: t\geq 0 \}$$. Since for any $$x,y \in X, x, we have $$[x,y] \neq [0,+ \infty)$$ because there exists $$|y|+1 \in [0,+\infty)$$ with $$|y|+1$$ not in $$[x,y]$$. Hence $$[0,+\infty)$$ is not a closed interval.
Well, is there any flaw in my understanding?
For reference, please see Topology by Munkres, page 84, chapter 2, section 14 the order topology:

page 85:

• According to the above definitions, $[0, +\infty)$ is not even an interval. It is a ray. – Gribouillis 2 days ago
• These are only details of terminology. Usually in ${\mathbb R}$, semi-infinite intervals are also called intervals. This doesn't seem to be the case in this book. Note that 'closed intervals' and 'closed sets' are defined separately. In the usual topology of ${\mathbb R}$, closed intervals turn out to be closed sets in the topological sense. – Gribouillis 2 days ago
• @Gribouillis Obviously we know that a ray is an interval, it's just an unbounded interval. I don't know why Munkres does not clarify that. IMO this is a flaw in his book, as it is clear that he ought to have done. – Prime Mover 2 days ago
• I think the definition varies from author to author. E.g. Rudin makes a distinction between sets of the form $(a,b)$, which he calls segments, and those of the form $[a,b]$, which he calls intervals. He calls $(a,b]$ and $[a,b)$ half-open intervals, and I don't even see a definition for $(-\infty,b)$ or $(a, \infty)$. But I think "most" authors, at least those writing analysis texts, would consider an interval to be any connected subset of $\mathbb R$, which would include both "intervals" and "rays" as defined by Munkres, as well as $(-\infty,+\infty) = \mathbb R$ and $(a,a) = \emptyset$. – Bungo 2 days ago
• @Bungo One could also perhaps take the definition $x, y\in I$ and $x\le z\le y$ imply $z\in I$. That would be an alternative to 'connected' using only the order relation. I don't know if some authors use this extended definition. – Gribouillis 2 days ago

Munkres is quite clear here: $$[0,+\infty)$$ is not a closed interval as it does not have an upper limit (the $$b$$ in the interval notation) in $$\Bbb R$$ but is a closed ray.

Both closed intervals and closed rays are closed sets in the order topology, as their name already suggests.

This is in $$\Bbb R$$

In the extended reals $$+\infty$$ is a point, so $$[0,+\infty)$$ is then a half-open interval, but it still contains the same points. It's no longer closed, as $$+\infty$$ is in its closure but not in it.