- #1
MathematicalPhysicist
Gold Member
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Well it's not homewrok cause i don't need to hand this question in, this is why i decided to put it here. (that, and there isn't a topology forum per se, perhaps it's suited to point set topology so the set theory forum may suit it).
Now to the question:
Show that if Y is a subspace of X, and A is a subset of Y, then the subspace topoology on A as a subspace of Y is the same as the subspace topology on A as a subspace of X.
(^-stands for intersection).
well so i wrote the specific topologies in hand:
T is a topology on X and T' a topology on Y.
T_Y subspace topology on Y from T, i.e: [tex]T_Y[/tex]={ Y^U|U\in T }
[tex]T_{A_Y}[/tex]={ A^U|U\in T' } is the subspace topology on A as a subspace of Y, and [tex]T_{A_X}[/tex]={ A^U| U\in T }
is the subspace on A as a subspace of X, now i need to prove that the latter topologies are the same, i.e are equal.
and so iv'e begun:
let V be in T_{A_Y} then there exists U in T' such that V=A^U but A is a subset of Y, then V is a subset of Y^U, so because Y and U are in T', also Y^U in T', now i need to prove that U is in T (the topology of X), now there's a lemma that ithink i need to use here but not certain how to apply it, the lemma states that:
Let Y be a subspace of X, if U is open in Y and Y is open in X then U is open in X.
now i obvioulsy want to show that U is open in X, so i need Y to be open in X, not sure how to do it, any hints?
p.s
also hints for the second part of the proof would be nice but i haven't yet tried to prove the second part.
Now to the question:
Show that if Y is a subspace of X, and A is a subset of Y, then the subspace topoology on A as a subspace of Y is the same as the subspace topology on A as a subspace of X.
(^-stands for intersection).
well so i wrote the specific topologies in hand:
T is a topology on X and T' a topology on Y.
T_Y subspace topology on Y from T, i.e: [tex]T_Y[/tex]={ Y^U|U\in T }
[tex]T_{A_Y}[/tex]={ A^U|U\in T' } is the subspace topology on A as a subspace of Y, and [tex]T_{A_X}[/tex]={ A^U| U\in T }
is the subspace on A as a subspace of X, now i need to prove that the latter topologies are the same, i.e are equal.
and so iv'e begun:
let V be in T_{A_Y} then there exists U in T' such that V=A^U but A is a subset of Y, then V is a subset of Y^U, so because Y and U are in T', also Y^U in T', now i need to prove that U is in T (the topology of X), now there's a lemma that ithink i need to use here but not certain how to apply it, the lemma states that:
Let Y be a subspace of X, if U is open in Y and Y is open in X then U is open in X.
now i obvioulsy want to show that U is open in X, so i need Y to be open in X, not sure how to do it, any hints?
p.s
also hints for the second part of the proof would be nice but i haven't yet tried to prove the second part.
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