1. Mar 19, 2006

### 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 dont know if my arguement 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 boundry 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 arguement 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?

2. Mar 19, 2006

### Muzza

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.

[0, inf) is not compact.

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.

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

It's indeed true.

3. Mar 19, 2006

### leon1127

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

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

because if the all boundary points isnt contained in such set, the set is said to be open.

Max and min exist imply that the lower and upper bounds exist and they are elements of that set. I cant think of any set that is open but contain max and min.

4. Mar 20, 2006

### Muzza

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"?

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".

5. Mar 20, 2006

### HallsofIvy

Staff Emeritus
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?

6. Mar 20, 2006

### leon1127

i figured it out. thx a lot!!