# An example showing subspace of compact space is not compact.

I had to give an example showing subspace of compact space is not compact. For this I first picked example of compact space from the book Topology by James R. Munkres section 26, solved example 2.

$$X= \{0\}\cup\{\frac{1}{n} | n\in \mathbb N \}$$. This space $$X$$ is compact space as subspace of $$\mathbb R$$. Now Let, $$Y= \{\frac{1}{n}\mid n\in \mathbb N\}$$ I claim that $$Y$$ is not compact as subspace of $$X$$. To show this I used the fact that if $$Y$$ is subspace of $$X$$ then $$Y$$ is compact iff every covering of $$Y$$ by sets open in $$X$$ contains a finite subcollection covering $$Y$$. First I showed that each singleton set $$\{\frac{1}{k} \}$$ is open in $$X$$. Then, $$Y=\bigcup_{k=1}^{\infty} \{\frac{1}{k} \}$$ So, this collection is open cover of $$Y$$ by sets open in $$X$$. Hence, by above fact if $$Y$$ is compact then there exist a finite subcollection of above singleton sets. But, this is not possible because, if such finite subcollection exist then $$Y$$ would be finite but, $$Y$$ is infact infinite set. So, we conclude that $$Y$$ is not compact set.

Is my solution is correct? If not, then please tell me corrections.

• Sure, that works. Jun 5 '20 at 8:33

In general, as long as you pick a closed and bounded subset of $$\mathbb{R}^n$$ (such as a closed ball), and take a non-closed subset (e.g open ball), it satisfies the requirements of the question. Your example is similar.