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# Interior[(interior(A)]=interior(A) Watch

1. Hello everyone I am trying to prove

Interior[interior(A)]=interior(A)

Where A is a set contained in R^n space,

If if start with proving the Int(A) subset Int[int(A)] then this is obvious because

If we let xEint(A), then as Int(A) is open and so there exist r>0 such that open B(x,r) is contained in int(A) and thus x is contained in Int[int(A)] so Int(A) subset Int[int(A)]

but how do I go about proving Int(Int(A)) subset int(A)?

Thank you, any help would be most appreciated!
2. I'm not sure what definition you're using for 'interior', but the one that I remember usually using was 'int A = the largest open set contained in A'. If we go with that definition then the proof almost writes itself. What definition are you working with?
3. (Original post by ljfrugn)
I'm not sure what definition you're using for 'interior', but the one that I remember usually using was 'int A = the largest open set contained in A'. If we go with that definition then the proof almost writes itself. What definition are you working with?
The union of all open sets contained in A is the interior of A, which is the largest open set contained in A. Yeah I know the statement is very obvious, but how do you prove it for n space?
4. (Original post by zcomputer5)
The union of all open sets contained in A is the interior of A, which is the largest open set contained in A. Yeah I know the statement is very obvious, but how do you prove it for n space?
It really is obvious isn't it? I mean, int(A) is the largest open set, so obviously since int(A) is already an open set, int(int(A)) is just int(A) by the definition of interior...
5. (Original post by zcomputer5)
The union of all open sets contained in A is the interior of A, which is the largest open set contained in A. Yeah I know the statement is very obvious, but how do you prove it for n space?
When unsure just use your sets;

The interior is the union of all open sets in A. Hence open. therefore any point in the interior is an interior point, viz. some neighbourhood exists around that point.

Hence

int is a subset of int(int)

Yet the interior of some arbitary set must be contained in that set

hence

int(int) is a subset of int

It is usually easy to argue with inclusion because you are only attempting to do a single thing at once - namely show inclusion in one specific direction; the form of this argument is much better from the way it rests explicitly on quantifiers (almost) so can be easily adapted to prove

The closure of the closure of a set is the closure i.e.

clos(closA) = clos(A)

a result that you could perhaps prove to accompany the proof above they are intimately related.
6. (Original post by DeanK22)
When unsure just use your sets;

The interior is the union of all open sets in A. Hence open. therefore any point in the interior is an interior point, viz. some neighbourhood exists around that point.

Hence

int is a subset of int(int)

Yet the interior of some arbitary set must be contained in that set

hence

int(int) is a subset of int

It is usually easy to argue with inclusion because you are only attempting to do a single thing at once - namely show inclusion in one specific direction; the form of this argument is much better from the way it rests explicitly on quantifiers (almost) so can be easily adapted to prove

The closure of the closure of a set is the closure i.e.

clos(closA) = clos(A)

a result that you could perhaps prove to accompany the proof above they are intimately related.
Thanks.

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