# Which of the following sets are compact:

Which of the following sets are compact:

1. $\{(x,y,z)\in \Bbb R^3:x^2+y^2+z^2=1\}$ in the Euclidean topology.
2. $\{(z_1,z_2,z_3)\in \Bbb C^3:{z_1}^2+{z_2}^2+{z_3}^2=1\}$ in the Euclidean topology.
3. $\prod_{n=1}^\infty A_n$ with the product topology where $A_n=\{0,1\}$ has discrete topology.
4. $\{z\in \Bbb C:|\operatorname{Re} z |\leq a \}$ for some fixed positive real number $a$ in the Euclidean topology.

$1$ is closed and bounded and hence compact,$2$ is closed but not bounded and hence not compact.

$3$ is compact by Tychonoff Theorem and $4$ is not bounded and hence not compact.

Are these correct?

• It seems that $1=2,$ but you say $1$ is compact while $2$ is not compact ? – Balloon Dec 23 '15 at 14:07
• The set in 2 is not compact if you change $\mathbb{R}^3$ into $\mathbb{C}^3$. Is it a typo? – egreg Dec 23 '15 at 14:09
• @Baloown;I have done the edits – Learnmore Dec 23 '15 at 14:13
• Now you are right ! – Balloon Dec 23 '15 at 14:14

That's correct. However, 1, 2 and 4 need a proof.

All three sets are closed, being inverse images of a closed set under a continuous function.

The set in 1 is bounded, because it is contained in $[-1,1]^3$.

The sets in 2 and 4 are not bounded, because they contain element with arbitrarily large norm; can you show them?

Set 4:

Easy: you can take $z=a+bi$ with arbitrary $b$.

Set 2:

Consider $z_3=1$. Then you can take $z_2=iz_1$, for arbitrary $z_1$.

• yes I can show them – Learnmore Dec 23 '15 at 14:34
• @Amartya I added the spoilers – egreg Dec 23 '15 at 14:46
• Yes I have got the counterexamples too ,thank you – Learnmore Dec 23 '15 at 14:57
• @Learnmore 1,2,4, I could prove myself. 3rd one it is beyond my imagination. How do I attack the (3) product topology related questions? please help me. – user464147 Oct 16 '17 at 10:56
• @N.Maneesh A product of compact spaces is compact. – egreg Oct 16 '17 at 11:41