Open cover and Finite Subcover

[michonamona]

Homework Statement

Show that each subset of R is not compact by describing an open cover for it that has no finite subcover.

b.) N (natural numbers)
The correct solution for this is A_n = (n-1/3, n+1/3) for all n in N. But the answer I came up with is [1,n) for all n in N. Is my answer also correct? For each x in N, there will always be a z in N such that z>x, so therefore the set [1,n) will not have a finite subcover for N.

d.) {x in Q: 0<=x<=2}
Correct solution:
A_0 = (sqrt(2), 3)
A_n = (-1, sqrt(2) - 1/n) for all n in N

I just want to confirm if my understand of this solution is correct. The reason why this works is because 1/n approaches 0 as n approaches infinity, but it never touches 0. Therefore, there will always be a space between sqrt(2)-1/n and sqrt(2). Thus, by density of the real number line, there will be a rational number in that space. Hence, we can never have a finite subcover.

I appreciate your patience. Thank you.

M

Homework Equations

The Attempt at a Solution

 

[lanedance]

is [1,n) open?

 

[lanedance]

also there's no one correct solution to these problems... your reasoning for the 2nd seems good to me

 

[michonamona]

is [1,n) open?

Ah...my mistake. How about (1,n) for all n in N? Is this an open cover of N with no finite subcover for N?

also there's no one correct solution to these problems... your reasoning for the 2nd seems good to me

Thank you.