compactness of sets in euclidean topology and product topology Which sets are compact in euclidean topology and product topology ?


*

*$\{(z_1,z_2,z_3):z_1^2+z_2^2+z_3^2=1)\}$ in the euclidean topology.

*$\{z\in \mathbb C:|re(z)|\leq a\}$ in the euclidean topology for some fixed positive real number $a$.

*$\prod A_n$ with product topology , where $A_n=\{0,1\}$ has discrete topology for n=0,1,........
I think (1) is not bounded in $\mathbb C$ ,so not compact . Are (2) and (3) compact or not? someone help. 
Thanks.
 A: $1.$ The set is not bounded since $(1,n,ni)$ is inside 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.
$2.$ The set is not bounded since $a+in$ is inside for every $n\in\mathbb{N}$ and has norm equal to $\sqrt{a^2+n^2}$ which goes to $+\infty$ as $n$ grows to $+\infty$. Hence the set is not compact.
$3.$ First, notice that $\{0,1\}$ is compact for the discrete topology. It his indeed Hausdorff, since $0$ and $1$ admits disjoint open neighborhoods, namely $\{0\}$ and $\{1\}$. Besides, let $(U_i)_{i\in I}$ be an open-cover of $\{0,1\}$, for every $i\in I$, one has $U_i\in\{\{0\},\{1\},\{0,1\}\}$ and since $\bigcup\limits_{i\in I}U_i\supseteq\{0,1\}$, there exists  $(i,j)\in I^2$ such that $U_i\cup U_j=\{0,1\}$, in other words $(U_i,U_j)$ is a finite open-subcover of $(U_i)_{i\in I}$. Otherwise, one gets: $$\bigcup\limits_{i\in I}U_i=\{0\}\not\supseteq\{0,1\}\textrm{ or }\bigcup\limits_{i\in I}U_i=\{1\}\not\supseteq\{0,1\}.$$ Hence, every open-cover of $\{0,1\}$ admits a finite open-subcover. 
Using Tykhonov's theorem, $\{0,1\}^{\mathbb{N}}$ is compact for the product topology.
Reminders: 


*

*If $X$ is equipped with discrete topology, then every subset of $X$ is both open and close.

*If $X$ is a topological space, $A\subseteq X$ is compact if and only if the topology induced by $X$ on $A$ makes $A$ be a Hausdorff space and from every open-cover of $A$ one can extract a finite open-subcover.
Remark. Have a look to your definition of a compact space, since it appears some people does not ask compact space to be Hausdorff.
