Closed interval is covering compact

[Scousergirl]

The question asks to prove directly that the closed interval is covering compact

- U= an open covering of the closed set [a,b]
I started by taking C=the set of elements in the interval that finitely many members of U cover. Now I need to somehow use the least upper bound theorem to show that b is in C?

 

[ZioX]

You want to show that for any arbitrary open cover of [a,b] there is a finite subcover.

Is covering compact the same thing as compact?

 

[Scousergirl]

Right so I assume U is my arbitrary open cover.

Covering compact implies compact but not vice versa right?

 

[morphism]

What are your definitions of compact and covering compact?

 

[Scousergirl]

for any arbitrary open cover of [a,b] there is a finite subcover for covering compact but I the question isn't asking to use the definition of compact just covering compact.

 

[matt grime]

That is the definition of compact most people use. You've been given slightly non-standard notation. This is also called quasi-compact by some as well.

You're on the right lines. Show that the set of points you described (as having the finite covering property) is closed, then think again.

 

[aday]

The question asks to prove directly that the closed interval is covering compact

- U= an open covering of the closed set [a,b]
I started by taking C=the set of elements in the interval that finitely many members of U cover. Now I need to somehow use the least upper bound theorem to show that b is in C? ineed solution

 

[HallsofIvy]

What makes you think that you can show that without showing, first, that there exist a finite subcover for the entire interval. That is NOT the usual proof for this statement.

The usual proof is by contradiction. Suppose there is NOT finite subcover for the interval. Now look at the two subintervals, [a, c], [c, b] where c is between a and b (for simplicity, you can choose it half way between- c= (a+b)/2. Since the entire interval cannot be covered by a finite collection of these open sets, at least one of the two subintervals cannot. Cut that interval into two pieces and repeat. Repeat until you get to a single point.

Of course, you have to be able to show that you do, in the limit, get a single point. For that you must specify that your interval is an interval or real numbers so that you can use the "least upper bound" and "greatest lower bound" properties. If you are thinking of [a, b] as an interval of rational numbers, the statement is not true.