1)show that a topological space X is metrizable iff there exists a homeomorphism

Question

1)show that a topological space X is metrizable iff there exists a homeomorphism of X onto a subspace of some metric space Y

2)Let X be a non-empty set, and consider the class of subsets of X consisting of the empty set and all sets whose complements are countable.Is this a topology on X?

Explanation / Answer

1) Let Y be a metric space and f:X->Y and f is a homeomorphism there exists a homeomorphism onto subspace of some metric space Y -> X is metrizable That means that there exists a dy which is a distance on Y then we simply take the image of the func dy under f ie dx(a,b) = dy(f(a),f(b)) its easy to see that this function is a distance. Now the other way X is metrizable ->there exists a homeomorphism onto subspace of some metric space Y Now X is a metric space so the identity homeomorphism is the required homeomorphism. 2)Yes this is in fact a topology called the countable complement topology. We need to prove that arbitrary unions and finite intersections of open sets are open. ie let A be a collection of open sets. 1. A is arbitrary collection. we will prove that union of the elements in A also is a open set. To see : Consider a in A, X - a is countable. Hence Union(a) = X - Intersection(X-a) [over a in A] but each of X-a is countable so thier intersection is also countable, hence X - Union(a) = Intersection(X-a) is also countable Hence arbitrary unions are open Similar proof for the finite intersections using the demorgans law. Message me if you have any doubts

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