Consider the following subsets of $\mathbb{R}^2$,where $a,b\in\mathbb{R}$

Question

consider the following subset of $\Bbb R^2$,where $a,b\in\Bbb R$:

$A=\{(x,y):\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,a\neq b\}$
$B=\{(x,y):\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leq1,a\neq b\}$
$C=\{(x,y):ax+by+5=0\}$
$D=\{(x,y):ax=by^2\}$
$E=\{(x,y):x^2+y^2=1\}$

Then which of the following is correct?

$C$ and $D$ are compact,but $A,B,E$ are not compact.
$A$ and $B$ are compact,but $C,D,E,$ are not compact.
$A,B$ and $E$ are compact but $C,D$ are not compact.
$A$ and $E$ are compact but $B,C,D$ are not compact.

How we can solve this

What are your thoughts up to now? I've reformatted the question; for your a#b I guessed $a\ne b$. — egreg, Jul 17 '13 at 10:59
In finite dimensional spaces, s set is compact iff it is closed and bounded. — Mhenni Benghorbal, Jul 17 '13 at 11:08

score 2 · Answer 1 · answered Jul 17 '13 at 11:11

Hint:

A Boundary set is a closed set, so $A, E$ is closed and since they are subset of $\mathbb R^2$ so they are both compact. The same story is true for B. In fact, it depicts a closed set including an interior set $\frac{x^2}{a^{2}}+\frac{y^2}{b^2}<1$ and a boundary set $\frac{x^2}{a^{2}}+\frac{y^2}{b^2}=1$.

score 1 · Answer 2 · answered Jul 17 '13 at 11:44

If $f\colon\mathbb{R}^2\to\mathbb{R}$ is continuous, then the inverse image of any closed subset is closed.

In particular, if $f(x,y)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$, you get that the inverse image of $\{1\}$ and of $[0,1]$ are both closed in $\mathbb{R}^2$. Therefore your sets $A$, $B$ and $E$ are closed (the last one is the case $a=b=1$).

Similarly, considering $g(x,y)=ax+by+5$, the set $C$ is the inverse image of $\{0\}$, so $C$ is closed. Also $D$ is closed, because it is the inverse image of $\{0\}$ under the function $h(x,y)=ax-by^2$.

Since you're interested in compactness, the only thing you have to check for the five sets is their boundedness or unboundedness.

Can you see which ones are bounded and which ones aren't?

score 0 · Answer 3 · answered Jul 17 '13 at 11:10

0

Correct answer should be 3. Just apply Heine-Borel theorem and see which sets are closed as well as bounded.

answered Jul 17 '13 at 11:10

Samrat Mukhopadhyay

16,277

how we can check which set closed as well bounded – sam Jul 17 '13 at 11:14
Set A is the points on an ellipse, so must be closed and bounded. Similarly check for Sets B (region inside the ellipse containing boundary) and E (circle). But set C is a straight line and set D is a parabola, so they are not bounded. – Samrat Mukhopadhyay Jul 17 '13 at 11:17
@santoshkumar: For showing that a set is closed, we should check every accumulate points of the set is also in the set, regarding the main topology. – Mikasa Jul 17 '13 at 11:18
I think the topology here is $\mathbb{R}^2$. Then I think, one can easily check the closedness of the sets in A,B, and E. – Samrat Mukhopadhyay Jul 17 '13 at 11:26
@sam: Another useful link – Mikasa Jul 17 '13 at 11:29
yes u are right samrat] – sam Jul 17 '13 at 11:38

Consider the following subsets of $\mathbb{R}^2$,where $a,b\in\mathbb{R}$

3 Answers3