Proving Equality of Subspace Topologies: A Topological Lemma Approach

In summary, the conversation discusses the proof that if Y is a subspace of X and A is a subset of Y, then the subspace topology on A as a subspace of Y is the same as the subspace topology on A as a subspace of X. The conversation goes into detail about the specific topologies and how to prove their equality using open sets and intersections. Ultimately, it is shown that the two topologies coincide, meaning they are the same.
  • #1
MathematicalPhysicist
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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.
 
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  • #2
loop quantum gravity said:
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.

Suppose O is open in the topology on A induced from Y. Then 0 = A n V for some V open in Y. Since V is open Y we have V = Y n B for some B open in X. Then 0 = A n V = A n Y n B = A n B because A is a subset of Y , hence O is open in the topology induced on A from X.

Conversely, suppose O is open in the subspace topology on A induced from X. Then O = A n B for some open B in X. Since A is a subset of Y, O = A n B = A n (Y n B) and Y n B is open in Y since B is open in X. Thus O is open in the topology induced on A from Y.

So the open sets in the topology induced on A from Y are exactly the same as the open sets in the topology induced on A from X. So these topologies coincide.

Notice two things. I didn't use any theorems, only what it means to be open in each respective topology, and, I didn't introduce any symbols for the topologies, which makes things easier and cleaner imho. Hope this helps.
 
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  • #3
I don't quite understand.
why because V is open in Y we have V=YnB for some B open in X, the fact that it's open in the subspace topology means that it's open in every other topology on Y, cause i think that this is what is assumed here, cause the above assertion is valid when V is in T_Y although from our assumption it's in T'.
 
  • #4
U is open in the subspace A of Y
iff U=A[itex]\cap[/itex]V where V is open in the subspace Y of X
iff U=A[itex]\cap[/itex](Y[itex]\cap[/itex]G)=A[itex]\cap[/itex]G where G is open in X (the second equality holds because A[itex]\cap[/itex]Y=A).
 
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  • #5
OK morphism repeating it doesn't explain it.

as I said U is in T_A_Y iff there exists V in T' such that U=AnV where T' is the topology on Y, the definition of subspace topology is quite clear every intersection of an open set in the topology on Y with A is in this topology.
now, if V is in T' does it means that it's in T_Y?
I mean does a set has only one topology on it ( i think not).
 
  • #6
It's stated that the topology on Y is the subspace topology it gets from X...
 
  • #7
you mean that by saying Y is a subspace of X, it's actually saying that the subspace topology is the topology on Y.
yes, I see your point, thanks.
 

1. What is subspace topology?

Subspace topology is a mathematical concept that describes a topology or a set of open sets on a subset of a topological space. It is used to study the properties of a subset and how it relates to the larger space it is a part of.

2. How is subspace topology different from the standard topology?

Subspace topology differs from the standard topology in that it is defined on a subset of a larger topological space, while the standard topology is defined on the entire space. This means that the open sets in the subspace topology are the intersections of the open sets in the larger space with the subset.

3. What are some applications of subspace topology?

Subspace topology has applications in many areas of mathematics, including topology, geometry, and analysis. It is also used in physics to study the properties of manifolds and in computer science for data analysis and machine learning algorithms.

4. How is the subspace topology related to the concept of continuity?

The subspace topology is closely related to the concept of continuity. If a function is continuous in the larger topological space, then it will also be continuous in the subspace topology. This means that the properties of continuity and convergence can be studied in the context of subspace topology.

5. Can different subsets of the same topological space have different subspace topologies?

Yes, different subsets of the same topological space can have different subspace topologies. This is because the subspace topology is defined by the open sets in the larger space, which can vary depending on the subset. Therefore, the subspace topology of one subset may not be the same as the subspace topology of another.

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