Questions on locally compact space

[jtceleron]

Q1. if A is a subset of X, choose the topology on X as {∅,U|for every U in X that A is a subset of U}. Then is this topology a locally compact space?
Q2. X=[-1,1], the topology on X is {∅, X, [-1, b), (a,b), (a,1] | for all a<0<b}. How to prove every open set (a,b) in X is NOT locally compact?

for Q1, there is so few restrictions on X, I don't know whether it's a locally compact space or not, however, I also cannot find a counterexample.
for Q2, when defined open set of a specific topology, does is mean that we have also defined closed set? If so, then for every point a in X, a closed set including a is a compact subset, then it is locally compact, a contradiction.

 

[Bacle2]

Usually, you find a condition satisfied by your space that is not satisfied by locally-compact spaces.

Example : Rationals as a subspace of Reals are not locally compact; in metric spaces,

compactness is equivalent to every sequence having a convergent subsequence.

Then, e.g. the sequences given by:

1) a₁=1, a₁=1.4 , a₃=1.41 ,... (first n terms of the decimal expansion of √2 )

2) a₁=1 , a₂=2,..., a_n =n ,...

are sequences without convergent subsequences. In non-metric spaces it is a little harder.

Still, re #2 : I have never seen a description of local compactness in terms of open sets,

but in terms of points. What definition are you using?

 

[jtceleron]

Usually, you find a condition satisfied by your space that is not satisfied by locally-compact spaces.

Example : Rationals as a subspace of Reals are not locally compact; in metric spaces,

compactness is equivalent to every sequence having a convergent subsequence.

Then, e.g. the sequences given by:

1) a₁=1, a₁=1.4 , a₃=1.41 ,... (first n terms of the decimal expansion of √2 )

2) a₁=1 , a₂=2,..., a_n =n ,...

are sequences without convergent subsequences. In non-metric spaces it is a little harder.

Still, re #2 : I have never seen a description of local compactness in terms of open sets,

but in terms of points. What definition are you using?

I mean, in #2, that if every point in the open set (a,b) has a compact subset, then the open set is locally compact.

 

[jtceleron]

in #1. the topology on X has already defined, which mean if choosing rationals as the subset A, then U are all the subsets including A, it is like a discrete topology, so imposing a metric topology on it seems not right.

 

[blue_raver22]

For Q1, the topology defined on X is not a locally compact space. To prove this, we can take the subset A = {0} and consider the open set U = (-1, 1). Since A is a subset of U, U is in the topology on X. However, there is no compact subset of X that contains U. This violates the definition of a locally compact space, which states that for every open set U in the topology, there exists a compact subset K of X such that U is a subset of K.

For Q2, we can prove that every open set (a, b) in X is not locally compact by contradiction. Assume that (a, b) is locally compact. This means that there exists a compact subset K of X such that (a, b) is a subset of K. However, since (a, b) is an open set, it cannot contain its boundary points a and b. This means that K must be a proper subset of X, which contradicts the fact that K is compact. Therefore, (a, b) cannot be locally compact.