Question

Prove that the union of two compact sets is compact using the fact that every open cover has a finite subcover.

Answer #1

show that each subset of R is not compact by describing an open
cover for it that has no finite subcover . [1,3) , also explain a
little bit of finite subcover, what does it mean like a finite.

Prove the following:
The intersection of two open sets is compact if and only if it
is empty. Can the intersection of an infinite collection of open
sets be a non-empty compact set?

(2)
If K is a subset of (X,d), show that K is compact if and only if
every cover of K by relatively open subsets of K has a finite
subcover.

Let X be a topological space with topology T = P(X). Prove that
X is finite if and only if X is compact. (Note: You may assume you
proved that if ∣X∣ = n, then ∣P(X)∣ = 2 n in homework 2, problem 2
and simply reference this. Hint: Ô⇒ follows from the fact that if X
is finite, T is also finite (why?). Therefore every open cover is
already finite. For the reverse direction, consider the
contrapositive. Suppose X...

For each of the following subsets of the real line, mark which
ones are compact, meaning that every open cover has a finite
subcover.
[1, infinity)
(1, 3)
[1, 3]
(1, 3]
{1, 3}
Just answer the question by choosing the correct answer
choices.

Use the fact that “countable union of disjoint countable sets is
countable" to prove “the set of all polynomials with rational
coefficients must be countable.”

Prove the union of a finite collection of countable
sets is countable.

Prove the union of two infinite countable sets is countable.

Prove that the union of infinitely many sets is countable using
induction.

Prove whether or not the intersection and union of two
uncountable sets must be uncountable

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