The metric space (0,1) is closed and bounded, but not compact?

In my lecture notes for one of my modules it states this, but I don't understand how it is closed and bounded, since it's an open interval. The metric space is the open interval (0,1) with the metric that comes as a subset of $$\mathbb{R}$$ .

This comes down to whether it's considered as a subset of $$\Bbb R$$, or as its own metric space.

As a subset of $$\Bbb R$$ it's clearly not closed, as it doesn't contain the end points $$0$$ and $$1$$. However, as a metric space on its own, it's closed because, well, all spaces are closed as subsets of themselves.

Clearly it's bounded (in fact totally bounded) as the distance between any two points is less than $$1$$. On the other hand, it's not complete, so it is not compact.

"Closed" is not a property of a topological space. It makes no sense to ask "Is the metric space $$X$$ closed?"

Instead, "closed" is a property of a subset of a topological space. Given a topological space $$X$$ and a subset $$A \subset X$$, we say $$A$$ is closed in $$X$$ is $$X \setminus A$$ is open in $$X$$.

It's very easy to abbreviate the phrase "$$A$$ is closed in $$X$$" to just "$$A$$ is closed". And if you keep very, very, very clear in your head what topological space $$X$$ the set $$A$$ is a subset of, then there's no problem.

The trouble starts (and it's a lot of trouble) when one forgets what space $$A$$ is a subset of.

If you had asked "Is $$(0,1)$$ a closed subset of $$(0,1)$$?" then there's no ambiguity: the answer is "Yes" because $$X = A = (0,1)$$ and every topological space is a closed subset of itself.