Are there specific conditions for functions in a non-documented metric?

[leon1127]

i need to mark statements below true or false and justify.
a) every nonempty finite set is compact.
since a finite set must have upper and lower bounds, thus it must be bounded. thus my question becomes that if a finite set is closed. i am considering a closed set [0,5] subset of R finite or not. it seems like it is not infinite set thus it is not compact. but i don't know if my argument is valid or not.
b) no infinite set is compact.
false, counterexample is [0,inf) subset of R
c)the union of any collection of open sets is an open set.
true because it is the size of such set shall only increase and the boundary point of the union set never conatined in the set.
d)if a set has a max and a min, then it is compact.
max and min imply it is bounded and closed, thus it is a compact set.
e) if a nonempty set is compact, then it has a max and min.
similar argument as d since the inverse is true.


and the last question was that find an example of a metric that is not shown in the book.
i use riemann integral[ |f-g|] and shown 4 properties of metric. what conditions do i need to put on functions f and g?
i said they are from R^n to R and continous. Do i need to state that they are increasing/decreasing monotonically too?


thx for any advice

 

[Muzza]

a) every nonempty finite set is compact.
since a finite set must have upper and lower bounds, thus it must be bounded. thus my question becomes that if a finite set is closed. i am considering a closed set [0,5] subset of R finite or not. it seems like it is not infinite set thus it is not compact. but i don't know if my argument is valid or not.

What IS your argument? To show that a finite set is closed, simply look at the definition. How is [0, 5] not infinite? But then again, I fail to see the relevance of [0, 5] at all for this question.

b) no infinite set is compact.
false, counterexample is [0,inf) subset of R

[0, inf) is not compact.

c)the union of any collection of open sets is an open set.
true because it is the size of such set shall only increase and the boundary point of the union set never conatined in the set.

I can't make sense of this at all. How are boundary points related to this question? I encourage you to simply look at the definitions.

d)if a set has a max and a min, then it is compact.
max and min imply it is bounded and closed, thus it is a compact set.

Having a maximum/minimum most certainly does NOT imply that a set is closed.

e) if a nonempty set is compact, then it has a max and min.
similar argument as d since the inverse is true.

It's indeed true.

 

[leon1127]

What IS your argument? To show that a finite set is closed, simply look at the definition. How is [0, 5] not infinite? But then again, I fail to see the relevance of [0, 5] at all for this question.

i mean i think [0,5] is not finite since it is uncountable.

counterexample for statement 2 is also [0,5].

I can't make sense of this at all. How are boundary points related to this question? I encourage you to simply look at the definitions.

because if the all boundary points isn't contained in such set, the set is said to be open.

Having a maximum/minimum most certainly does NOT imply that a set is closed.

Max and min exist imply that the lower and upper bounds exist and they are elements of that set. I can't think of any set that is open but contain max and min.

 

[Muzza]

because if the all boundary points isn't contained in such set, the set is said to be open.

I don't think this definition (characterization) of open sets is the easiest one to work with in this case. What about "a set M is open iff there exists a neighbourhood X around each point x in the set such that X is contained in M"?

Max and min exist imply that the lower and upper bounds exist and they are elements of that set. I can't think of any set that is open but contain max and min.

You don't need to find an open set which has a max and min, you need to find a set which is not closed and has a max and min. Note that "not closed" is not the same as "open".

Think about unions of intervals...

 

[HallsofIvy]

It is true that a closed and bounded set in Rⁿ is compact.

However, it true that a finite set in any topology is compact- that follows immediately from the definition of compact as "every open cover contains a finite subcoverf". What definition of compact are you using?

 

[leon1127]

i figured it out. thanks a lot!