Help with Topology: Density and Customly Defined Similarity

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Homework Statement


The problem is as follows:
Two topological space, T1 and T2, similar, T1~T2, iff for every non-empty V1 in T1 there is a non-empty V2 in T2 s.t. V2<V1 and for every non-empty V2 in T2 there is a non-empty V1 in T1 s.t. V1<V2.
Prove (a) and (b):
a) If T1 and T2 have the same dense sets, they are similar.
b) If T1~T2 and V is in T1 then the closure of V in T1 = the closure in T1 of the interior in T2 of V. ie. cl1(v)=cl1(int2(V)).

Homework Equations


We are using the definition that D is dense if its closure is the whole space. Proofs using this definitions would be best as we were not taught any equivalent ones.
Not sure if relevant but just in case:
~ is an equivalence relation; ~ topologies share dense sets; ~ topologies share sets with the empty interior

The Attempt at a Solution


I started (a) by saying T1 and T2 share dense sets and assuming they aren't ~ but I have yet to see anyway to reach a contradiction. I also thought about trying a contra-position but I'm not sure it would pay off either. I'm pretty clueless about how to approach (b) altogether. I would appreciate any help you can offer even if it just get me started. These are actual the last questions in a set of questions and I already managed to get the earlier ones these I just can't see. Thanks for any help.
 

Discussion

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Okay so now I would have, [itex]X\setminus cl_2(V_1)[/itex] which is an open set in [itex]\tau_2[/itex] is either contained in or contains [itex]X\setminus cl_1(V_1)[/itex] which is an open set in [itex]\tau_1[/itex]? This seems right because either the closure in T1 has to be bigger or the closure in T2 has to be but where does denseness come into play? I feel like I didn't go in the right direction here.
 
Okay, so [itex]V_1[/itex] is open in T1 implying that [itex]X\setminus V_1[/itex] is closed in T1. [itex]X\setminus V_1 \subset cl_2(X\setminus V_1)[/itex] and hence the closure in T2 must share some elements with V1. Thus [itex]X\setminus cl_2(X\setminus V_1) \subset V_1[/itex] and [itex]X\setminus cl_2(X\setminus V_1)[/itex] is open in T2. This seems solid but I don't see where to use the fact the space share the same dense sets and that seems like a really important part to not use. Also sorry about all the follow up questions I just want to make sure I get it.
 
skitzolala said:
Okay, so [itex]V_1[/itex] is open in T1 implying that [itex]X\setminus V_1[/itex] is closed in T1. [itex]X\setminus V_1 \subset cl_2(X\setminus V_1)[/itex] and hence the closure in T2 must share some elements with V1. Thus [itex]X\setminus cl_2(X\setminus V_1) \subset V_1[/itex] and [itex]X\setminus cl_2(X\setminus V_1)[/itex] is open in T2. This seems solid but I don't see where to use the fact the space share the same dense sets and that seems like a really important part to not use. Also sorry about all the follow up questions I just want to make sure I get it.

Yes. So your candidate for [itex]V_2[/itex] is [itex]X\setminus cl_2(X\setminus V_1[/itex]. But why is this set not empty??
 
I suppose you would assume that [itex]X\setminus V_1[/itex] is not dense hence the compliment of its closure can't be the empty set?
 
Since [itex]X\setminus V_1[/itex] is closed in T1 it would be it's own closure. Hence it isn't dense in T1 so it can't be dense in T2.