#Set theoretic problem (undergraduate level)

19 messages · Page 1 of 1 (latest)

rancid parrot
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I just wonder if my reasoing is correct. The problem is the following: "If A ⊆ B, then what is A ⋂ B, A \ B, and A ⋃ B?"

The way I understood is that since A is a subset of B, then every element x in A also exists in B. Since

A ⋂ B = {x : x ∈ A ∧ x ∈ B},

then it follows that A ⋂ B = A, because when you take the intersection of A and B, you are essentially finding the set of elements that exist in both A and B.

I applied the same reasoning to A \ B. Since

A \ B = {x : x ∈ A ∧ x ∉ B}

then it follows that A \ B = {}. In other words, the relative complement of B in A must not contain any elements. Since all elements x that exist in A also exist in B and x ∉ B, it cannot be the case that the relative complement B in A is an non-empty set.

Same thing with A ⋃ B. Since the definition of the union of A and B is as follows

A ⋃ B = {x : x ∈ A ∨ x ∈ B},

then the union is essentially the same as B because it already contains all elements in A and, possibly, additional elements.

I apologise for any grammar mistakes since English isn't my mother tounge, which is why I wonder if my reasoning is correct.

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crystal atlas
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Your reasoning is correct, but it isn't a good way to prove. Delete those words "essentially", "must not contain", "possibly"...

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It destroy your reasoning.

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But when you say like "A ⋂ B = {x : x ∈ A ∧ x ∈ B}", it's awesome. It's clear, it's definition.

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For example, instead of saying "when you take this, then take this, you transform this, and TADA", you could say :
Suppose that $A = A \cap B$.
Let us show that $A \subseteq A \cap B$ and then $A \cap B \subseteq A$. \

  • First, let $x \in A$, by hypothesis, $A \subseteq B$, so $x \in B$. That means $x \in A \cap B$. So $\forall x \in A, x \in A \cap B$. So $A \subseteq A \cap B$ \
  • Second, ...
inner girderBOT
crystal atlas
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That can only improve your math skills

rancid parrot
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Damn, proof writing is really something else. I'm just a first-year undergraudate student so my proof writing skill is weak. So based on your feedback, I should revise the second part like this:

Suppose A \ B = {}. Let us show that A \ B ⊆ ∅.

Let x ∈ A, by hypothesis, A ⊆ B, so x ∈ B. Then A \ B = {x : x ∈ A ∧ x ∉ B}. If A ⊆ B, then ∀x : [x ∈ A → x ∈ B]. That means A \ B = {}. So A \ B ⊆ ∅.

vagrant vapor
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The only thing I'd add is to write out the definition of A ⊆ B, just to make the reasoning more explicit.

We could write, say:

Supoose that A ⊆ B.
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Then, for any object x, it follows that x ∈ A → x ∈ B. Hence, the set S = {x : x ∈ A ∧ x ∈ B} is equal to the S' = {x : x ∈ A}.
_ _
But by definition S = A ∩ B and S' = A. Hence, A ∩ B = A.
_ _
Therefore, A ⊆ B → A ∩ B = A.

crystal atlas
rancid parrot
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@crystal atlas I just realised it

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I'll take it from here and improve my arguments

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Thanks @vagrant vapor @crystal atlas

sleek horizonBOT
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@rancid parrot has given 1 rep to @vagrant vapor @crystal atlas

rancid parrot
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+close