Is My Reasoning Correct for Determining Compactness?

[hartigan83]

I have been asked to determine if the following statements are ture or false...I have attepmted to answer each question using my understanding of the definitions of compact (a set is compact iff every open cover of S contains a finite subcover), the statement of the Heine-borel Theorem( A subset S of R is compact iff S is closed and bounded), and the Bolzano-Weierstrass theorem( If a bounded sub set S of R contains infinitely many points, then there exists at least one point in R that is an accumulation point of S). Please tell me if I am using reasoning that is not correct for any of them...Thanks

Every finite set is compact.
True. ex [0,3]

No infinite set is compact.
False, but I cannot think of a counter example...
I was thinking that a set could be bounded and still contain infinitley many points, but am not sure if it would be closed.

If a set is compact then it has a max and a min.
False it must also be non empty.

If a set has a max and a min then it is compact.
False, (2,3) has a max and a min but is not ompact.

Some undounded sets are compact.
False due to the hiene-borel theorem

If S is a compact subset of R then there is at least one point in R that is an accumulation point of S
Flase, it need not be compact, just bounded and contain infinitely many points.

If S is compact and x is an accumulation point of S then x is an element of S.
True, Since S is closed it contains all of its accumulation points.

If S is unbounded, then S has at least one accumulation point.
False, S must be bounded, and contain infinitely many points by Bolozano -weierstrass.

 

[Hurkyl]

Every finite set is compact.
True. ex [0,3]

Agreed, but why?

[0,3] is not an example: [0,3] is an infinite set.

No infinite set is compact.
False, but I cannot think of a counter example...
I was thinking that a set could be bounded and still contain infinitley many points, but am not sure if it would be closed.

Agreed. You already stated a counter-example.

Incidentally... if you're given any set S, there's an easy way to find a closed set that contains S...

If a set is compact then it has a max and a min.
False it must also be non empty.

Agree that it's false. You just need to state that the empty set is a counterexample.

That said, if you want to claim that any nonempty compact set has a max and a min, you need a reason!

If a set has a max and a min then it is compact.
False, (2,3) has a max and a min but is not ompact.

Agreed that the answer is false.

(2,3) is not a counter example, though: it has neither a max nor a min.

Some undounded sets are compact.
False due to the hiene-borel theorem

Agreed.

If S is a compact subset of R then there is at least one point in R that is an accumulation point of S
Flase, it need not be compact, just bounded and contain infinitely many points.

Answer is false, but
(1) your reasoning is wrong
(2) You don't need to reason, just produce a counter example.

The problem with your reasoning is that you are arguing against the converse! You sound like you're trying to disprove:

"If there is at least one point in R that is an accumulation point of S, then S is a compact subset of R".

If S is compact and x is an accumulation point of S then x is an element of S.
True, Since S is closed it contains all of its accumulation points.

Agreed.

If S is unbounded, then S has at least one accumulation point.
False, S must be bounded, and contain infinitely many points by Bolozano -weierstrass.

Agree that it's false. Your reasoning is wrong, though. You've made the same mistake as you did two questions ago. The B-W theorem says:

"If S is bounded and infinite, then S has an accumulation point"

In particular, it does not say:

"If S has an accumulation point, then S is bounded and infinite".

(In fact, it's easy to find unbounded sets that have accumulation points)

 

[hartigan83]

Agreed, but why?

[0,3] is not an example: [0,3] is an infinite set.

Incidentally... if you're given any set S, there's an easy way to find a closed set that contains S...

How?

Agree that it's false. You just need to state that the empty set is a counterexample.

That said, if you want to claim that any nonempty compact set has a max and a min, you need a reason!

I have a lemma in my book that states this.

Agreed that the answer is false.

(2,3) is not a counter example, though: it has neither a max nor a min.

I am not sure what to look at for a counter example. I was thinking the closed interval [2,3] but then that is compact so it does not work

Answer is false,...

[1,5] is compact and the set of accumulation points is [1,5] and 1 is an element of R

Agree that it's false. Your reasoning is wrong, though. You've made the same mistake as you did two questions ago. The B-W theorem says:

"If S is bounded and infinite, then S has an accumulation point"

In particular, it does not say:

"If S has an accumulation point, then S is bounded and infinite".

(In fact, it's easy to find unbounded sets that have accumulation points)

Are open intervals bounded?

 

[hartigan83]

Agreed, but why?

[0,3] is not an example: [0,3] is an infinite set.

What if I specified that it was the set of Natural numbers?

 

[Office_Shredder]

the set of natural numbers is infinite... a set like {1, 2, 3} is finite. Or perhaps just {1}