Which of the Following Sets are compact (C.S.I.R 2015)

Question

1. $\{ (x,y,z) \in \mathbb R^3 : x^2 + y^2 + z^2 = 1 \}$ in the Euclidean Topology

2. $\{ (z_1,z_2,z_3) \in \mathbb C^3 : z_1^2 + z_2^2 + z_3^2 = 1 \}$ in the Euclidean Topology.

3. $\prod_{n=1}^{\infty} A_n$ with product topology , where $A_n = \{0,1\}$ has discrete topology for $n = 1,2, \dots$

4. $\{ z\in \mathbb C : Re z \leq \alpha \}$ in the Euclidean topology for some fixed positive real number $\alpha$.

I think

For 1) and 2)

I know that any subset in $\mathbb R^n$ which is closed and bounded iff it is compact.

for 3)

Since $A_n$ is a finite discrete topological space . So it is compact. So by Tychnoff's Theorem $\prod A_n$ is compact.

for 4)

Since $K= \{ \alpha + i n : n \in \mathbb N \}$ is a subset of the given set which is unbounded. So it is not comact.

Please check my Solution, If you find any error , then correct me. Thank you

Answer 1

Your solutions seem correct to me. For the last one you could like to be able to show a direct contradiction of the definition of compactness, that is, an open cover which does not admit finite subcover.

This counterexample could be easily provided by a family of open subsets of $K$ such as $$U_n=K\cap\{z\in\mathbb{C} : n-1<Imz<n+1\}, ~ n\in\mathbb{N}$$

I guess no further explanation is required for the first three questions.

A note about question 3: $A_n$ is not required to be discrete in order to prove compactness. A finite set will always be compact, in any topology.

Answer 2

The second set is not compact, as the set is not bounded since $(1, n, ni)$ is an element for every $n\in \mathbb{N}$ and has norm equal to $\sqrt{1+2n^2}$ which goes to $\infty$ as $n$ grows to $\infty$. Hence, the set is not compact.