# Why the space X=(0,1) is (not sequentially) compact?

• I
My problem is that the space $X= (0,1)$ is not sequentially compact and compact at the same time.

It is not sequentially compact:
If we define the sequence $(\frac{1}{n})$ we can show that it is not sequentially compact as the sequence converges to 0, but $0 \notin X$.

It is compact:
On the other hand, for X to be compact we need
1) bounded: The space X is bounded as any ball with center $x \in X$ and radius 2 will X.
2) closed: Is closed as its complement is the empty set (which is open)

Thus, the set $X$ is compact, which is a contradiction as X is not sequentially compact.

Where is my mistake when I show that X is compact?

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Closed and bounded does not imply compact in general.

You are using the Heine-Borel theorem which only holds for subsets of Euclidean space.
In that case the interval isn't closed either.

Edit; Ninja'd
Edit 2; Subsets of Euclidean space as far as I know, there might be other special cases.

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I see, this works for subsets of $R^n$

lavinia
Gold Member
For a subset of a metric space, closed should mean that every Cauchy sequence converges to a point in the space. But a Cauchy sequence in the open unit interval that converges to 0 or 1 will not coverage to a point in the open interval. So the open unit interval is not closed in ##R^1##. If every Cauchy sequence in a metric space converges then the metric space is said to be complete.

Here is a metric space that is closed and bounded but not compact. On a closed disk in the plane, let the distance between two points be the Euclidean distance if the two points lie on the same radial line through the center of the disk; and the sum if their distances to the center of the disk if they lie on different radial lines.

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• suremarc
• JorisL
A follow-up question that's related to my research, does a similar theorem exist for a Riemannian submanifold of a Pseudo-Riemannian manifold?

• JorisL