2

For each set, determine if it is compact, and if not, why not(it's not closed or not bounded $?$)

$1.\{(x,y):1<e^{x^2+y^4}(x^4+y^2+1)<2\}$

$2. \{(x,y,z):|x|+|y|+|z|≤3\}$

$3. ⋃_{n∈\mathbb{Z}^+}[0,2−\frac{1}{n}]⊂\mathbb{R}$

$4. \{(x,e^x):0≤x≤10\}$

$5. \{(x,y):1≤e^{x^2+y^4}(x^4+y^2+1)≤2\}$

$6. \{(x,y):y=\sin(x)\}$

I checked the definition of compect set

Def. compect set

A set $S⊆\mathbb{R}^n$ is said to be compact if every sequence in $S$ has a subsequence that converges to a limit in $S$.

Seems like if the set isn't compact, then it's either not bounded or not closed,

Take the contrapositive have bounded and closed implies compact$\dots$

My attempts

$1.$ seems bounded but not closed so not compact

$2.$ closed and compact so compact

$3.$ union of closed set is closed and closed set are bounded so compact

$4.$ bounded so compact

$5.$ bounded so compact

$6.$ not bounded so not compact

Which might not be correct$\dots$

I'm a little confused about those concepts,

Any help would be appreciated.

Community
  • 1
Ethan
  • 5,291

1 Answers1

1

All of yous answers are correct other than (3). Since union of arbitrary closed sets need not be closed. $\cup_{n\in \mathbb{Z}^+}[0,2-\frac{1}{n}]=[0,2)$. It can not contain $2$, so not closed. Therefore it is not compact. For (2), (4),(5) and (6) you may use the result continuous image of compact set is compact and product of compact set is compact. For (1), it is not closed, it is cumbersome to show this, but by some calculation you can do that.

MANI
  • 1,928
  • 1
  • 11
  • 23