Proving Set Equality: A Simple and Effective Method

[Jairo Rojas]

Homework Statement

Attached is the problem

Homework Equations

The Attempt at a Solution

So I have to show that each side is a subset of the other side

Assume x∈ A ∪ (∩Bi)
so x∈A or x∈∩Bi

case 1 x∈ ∩ Bi

so x∈ (B1∩B2∩B3...∩Bn)
which implies x∈B1 and x∈B2 ... and x∈Bn
so x∈B1∪A and x∈B2∪A... and x∈Bn∪A
so x∈∩(A∪Bi)

My teacher told me that this approach doesn't work because the sets can be infinite and told me to use "words". I don't what he meant with that.

 

[haruspex]

My teacher told me that this approach doesn't work because the sets can be infinite and told me to use "words". I don't what he meant with that.

You don't need to use words instead of symbols, though you could. What you do need to do is avoid writing it as a finite (or even countably infinite) intersection. It is not hard to convert your proof to one that works.

 

[Jairo Rojas]

You don't need to use words instead of symbols, though you could. What you do need to do is avoid writing it as a finite (or even countably infinite) intersection. It is not hard to convert your proof to one that works.

can I say n->infinity?

 

[Jairo Rojas]

You don't need to use words instead of symbols, though you could. What you do need to do is avoid writing it as a finite (or even countably infinite) intersection. It is not hard to convert your proof to one that works.

what about if I split (∩Bi) into two finite sets call it S and B so it equals SnB

 

[andrewkirk]

The problem is that by writing x∈ (B1∩B2∩B3...∩Bn) you have assumed that the index set $I$ has only $n$ elements. But in fact it may be infinite, or worse, uncountable. That assumption cannot be made because it may be false. Consider for instance if the index set $I$ is the set of all real numbers in the interval [0,1) and $B_j$ for $j\in I$ is the set of all positive real numbers whose fractional part is $j$. Then $I$ is uncountably infinite.

To avoid making that invalid assumption, use the quantifier $\forall$, which means 'for all'.

So you have $x\in\bigcap_{j\in I} B_j$ and instead of writing

x∈B1 and x∈B2 ... and x∈Bn

you write $\forall j\in I:\ x\in B_j$.

Similarly for the rest of your proof: wherever you find yourself using $n$ or an ellipsis (that's the '...' you've written in the middle of lists), get rid of them by re-writing using $\forall$.

When you come to do similar proofs for unions of indexed sets you will need to use the other quantifier $\exists$, which means 'there exists' ('there is at least one').

 

[Jairo Rojas]

The problem is that by writing x∈ (B1∩B2∩B3...∩Bn) you have assumed that the index set $I$ has only $n$ elements. But in fact it may be infinite, or worse, uncountable. That assumption cannot be made because it may be false. Consider for instance if the index set $I$ is the set of all real numbers in the interval [0,1) and $B_j$ for $j\in I$ is the set of all positive real numbers whose fractional part is $j$. Then $I$ is uncountably infinite.

To avoid making that invalid assumption, use the quantifier $\forall$, which means 'for all'.

So you have $x\in\bigcap_{j\in I} B_j$ and instead of writing

x∈B1 and x∈B2 ... and x∈Bn

you write $\forall j\in I:\ x\in B_j$.

Similarly for the rest of your proof: wherever you find yourself using $n$ or an ellipsis (that's the '...' you've written in the middle of lists), get rid of them by re-writing using $\forall$.

When you come to do similar proofs for unions of indexed sets you will need to use the other quantifier $\exists$, which means 'there exists' ('there is at least one').

thanks!

 

[Jairo Rojas]

The problem is that by writing x∈ (B1∩B2∩B3...∩Bn) you have assumed that the index set $I$ has only $n$ elements. But in fact it may be infinite, or worse, uncountable. That assumption cannot be made because it may be false. Consider for instance if the index set $I$ is the set of all real numbers in the interval [0,1) and $B_j$ for $j\in I$ is the set of all positive real numbers whose fractional part is $j$. Then $I$ is uncountably infinite.

To avoid making that invalid assumption, use the quantifier $\forall$, which means 'for all'.

So you have $x\in\bigcap_{j\in I} B_j$ and instead of writing

x∈B1 and x∈B2 ... and x∈Bn

you write $\forall j\in I:\ x\in B_j$.

Similarly for the rest of your proof: wherever you find yourself using $n$ or an ellipsis (that's the '...' you've written in the middle of lists), get rid of them by re-writing using $\forall$.

When you come to do similar proofs for unions of indexed sets you will need to use the other quantifier $\exists$, which means 'there exists' ('there is at least one').

by the way I am theMathNoob. can you ask the administrator to unban my account?. I promise I won't post silly ps4 questions.

 

[berkeman]

by the way I am theMathNoob. can you ask the administrator to unban my account?. I promise I won't post silly ps4 questions.

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