- #1

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If I know that A is a subset of B,

what can I say about the relationship between A-complement and B?

what can I say about the relationship between A-complement and B?

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- Thread starter 1MileCrash
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- #1

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If I know that A is a subset of B,

what can I say about the relationship between A-complement and B?

what can I say about the relationship between A-complement and B?

- #2

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Draw a Venn diagram.

There IS no relationship that I can see.

There IS no relationship that I can see.

- #3

Office_Shredder

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- #4

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If [itex]A\subseteq B\subseteq X[/itex], then [itex]X\setminus A \supseteq X\setminus B[/itex].

- #5

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If [itex]A\subseteq B\subseteq X[/itex], then [itex]X\setminus A \supseteq X\setminus B[/itex].

That doesn't seem related to my question.

- #6

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Draw a Venn diagram.

There IS no relationship that I can see.

Ok then, my logic broke down severely while doing this proof, somewhere. I'll have to post about it soon, but I'm still too "in the fray" to do it now.

- #7

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Hey,

yeah, the proof is a little complex for my level and was probably dumb for me to attempt in the way I did, but basically my goal at this stage is that I'd have to show that:

[itex]A \cap B = \oslash[/itex]

and

[itex]A \subseteq B[/itex]

means precisely that

[itex]B \subseteq X \setminus A [/itex]

Where X is the universe, and that top thing is the null set, which I couldn't find.

I'm really confusing myself. Do you guys think that this implication is even true? If one of these conditions is not required then it has to be that I messed up somewhere.

EDIT writing it out made me see that these two first conditions are (almost) not compatible..

- #8

Office_Shredder

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From there of course you can prove that [itex] B \subset X\setminus A [/itex] but that's not a very interesting relationship anymore :tongue:

- #9

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From there of course you can prove that [itex] B \subset X\setminus A [/itex] but that's not a very interesting relationship anymore :tongue:

Right, so I think I see my problem.

At the start, I said that the closure of a set B is equal to the complement of the union of all open sets which are disjoint from B. I don't see any definition that agrees with that anywhere, I just thought it seemed reasonable.

- #10

Office_Shredder

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[tex] \bigcap_{i\in I} C_i [/tex]

where the C

[tex] = \bigcap_{i\in I} X\setminus U_i = X\setminus \bigcup_{i\in I} U_i [/tex]

which is exactly how you are calculating the closure - I think your statement is an accurate description of the closure of B.

You might be better served just starting a new thread with your whole proof to figure out where it goes wrong at this point.

- #11

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[tex] \bigcap_{i\in I} C_i [/tex]

where the C_{i}are the closed sets containing B. Taking complements, let [itex]U_i = X\setminus C_i [/itex] be the open sets which are disjoint from B. Then the closure of B is

[tex] = \bigcap_{i\in I} X\setminus U_i = X\setminus \bigcup_{i\in I} U_i [/tex]

which is exactly how you are calculating the closure - I think your statement is an accurate description of the closure of B.

You might be better served just starting a new thread with your whole proof to figure out where it goes wrong at this point.

Ok, thanks a lot for the help, I'll post tomorrow. I'm very interested to know what's going wrong!

- #12

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[tex] \bigcap_{i\in I} C_i [/tex]

where the C_{i}are the closed sets containing B. Taking complements, let [itex]U_i = X\setminus C_i [/itex] be the open sets which are disjoint from B. Then the closure of B is

[tex] = \bigcap_{i\in I} X\setminus U_i = X\setminus \bigcup_{i\in I} U_i [/tex]

which is exactly how you are calculating the closure - I think your statement is an accurate description of the closure of B.

You might be better served just starting a new thread with your whole proof to figure out where it goes wrong at this point.

So wait, looking again, isn't your first line the definition of the boundary of B, not the closure?

Never mind, my notes are wrong.

It all makes perfect sense now. I defined boundary and closure to be essentially the same. Then I have it being equal to itself remove a union. The result pointed out above shows that all sets of the union are empty, which is true for that silly equality to work!

Finding errors are pretty fun.

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