Set Theory Proof

[civic musk]

Prove that for any set $A$:
$$\emptyset \subseteq A \land A \subseteq A$$

[midnight halo]

Kenny

[civic musk]

Can anyone help with this problem? I am new to set theory and have no experience with writing proofs.

[midnight halo]

Tigre_Rampant

[oblique ruin]

first off there's an issue with how you parse the formula

$\land$ is something that combines two statements, $A \land A$ where $A$ is a set is meaningless

[midnight halo]

aPlatypus

[oblique ruin]

$(\emptyset \subseteq A) \land (A \subseteq A)$

[midnight halo]

aPlatypus

[oblique ruin]

this makes sense

i.e. you need to prove two things

• emptyset subseteq A

• A subseteq A

@civic musk

[civic musk]

How do I prove these two?

[oblique ruin]

oh I thought you were the one who wrote this thing @civic musk

oops

alright

well how do you understand \subseteq and \emptyset ?

what do they mean ?

@civic musk

[civic musk]

Subseteq means that the set on the left side is a subset or equal to the set on the right side.

Emptyset is a set with no elements.

[oblique ruin]

and what's a subset exactly ?

how would I check that A is a subset of B for example ?

[civic musk]

A set that consists of some or all of the elements of another set

A needs to have common elements with B

[oblique ruin]

that's a nice intuitive idea for a subset, but in math it's restated in a tiny bit different way

everything in A also has to be in B

in more math speak "for all x, if x is in A then x is in B"

[civic musk]

I am trying to learn how to think mathematically

[oblique ruin]

and with the math definition, this might be false

exactly because of the empty set

like if we took your definition, yeah emptyset subseteq A would be wrong

cause you require to actually have common elements btw A and B

[oblique ruin]

well here it's more remembering what math ppl mean when they say something, i.e. definitions

[civic musk]

True, I remember that emptyset is considered a subset of every set.

So for any set A, emptyset subseteq A should be always true

[oblique ruin]

or said another way "forall x in A, x is in B"

[oblique ruin]

you can think of that "forall x in A, ..." as a long ass list of requirements

with as many requirements as there are elements of A

but if A is empty there is no requirements to satisfy

[oblique ruin]

so the sentence is automatically true

[oblique ruin]

for a more concrete example, {1, 2, 3, 4} subseteq B, that means

• 1 in B
• 2 in B
• 3 in B
• 4 in B
  you require 4 things to be true about B

[civic musk]

Since the empty set has nothing, there's nothing to check

[oblique ruin]

yea

[civic musk]

And set A is a subset of A because anything that's inside A is still in A

I don't understand what more is there to prove

It looks obvious

How does one express all of this mathematically?

To be considered a valid proof

[oblique ruin]

for the second proof at least you can use more typical methods of proof, cause there's no way outside of explaining that quirkiness of the empty set in some way in the first one

the concept is called vacuous truth https://en.wikipedia.org/wiki/Vacuous_truth

[oblique ruin]

now for the second one, do you remember the typical way to show a statement of the form "for all x in A, ...." ?

[civic musk]

For all x in A, x is in A

[oblique ruin]

so yeah

pick some x in A, oh x is actually in A, done.

that's how the proof goes mathematically speaking too

are you using a book or something for studying set theory currently ?

@civic musk

[civic musk]

It's a Stanford MOOC called Introduction to Mathematical Thinking

[oblique ruin]

aight

well one book I like to recommend is the book of proof

it's free online, just google it

it helps you not get caught in routine proofs, it's pretty well done

@civic musk