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## Main Question or Discussion Point

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?

- Thread starter 1MileCrash
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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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Draw a Venn diagram.

There IS no relationship that I can see.

There IS no relationship that I can see.

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Office_Shredder

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

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That doesn't seem related to my question.If [itex]A\subseteq B\subseteq X[/itex], then [itex]X\setminus A \supseteq X\setminus B[/itex].

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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.Draw a Venn diagram.

There IS no relationship that I can see.

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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..

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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:

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Right, so I think I see my problem.

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:

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.

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

[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.

- #12

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

[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.

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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