Understanding some basic set theory stuff.

[U.Renko]

This is more to see if I understand it or not.
There are four statements and I need to explain why they are true. (they all are)
I understand it why some of they are, but my answers just don't feel accurate/formal enough.

Homework Equations

1) \mathbb{R}^3 \subseteq \mathbb{R}^3
2) \mathbb{R}^2 \nsubseteq \mathbb{R}^3
3) \left \{ (x,y): x - 1= 0 \right \} \subseteq \left \{ (x,y): x^2 - x = 0 \right \}
4) \left \{ (x,y): x^2 - x = 0 \right \} \nsubseteq \left \{ (x,y): x - 1 =0 \right \}

The Attempt at a Solution

1) is true simply because X is a subset of X for any set X. no problem with this one

2) is not so obvious for me.
I understand that one consists of ordered pairs and the other of ordered triples. But I'm not sure if this affects anything.

3) is true because { 1 } is a subset of { -1 , 1}

4) is true because { -1 , 1 } is not a subset of { 1 }


By The way: I was having trouble with 3 and 4. But I kinda got an insight while typing.
not sure if it could be more formal maybe.

So the really troubling one is 2.

 

[Simon Bridge]

1) is true simply because X is a subset of X for any set X. no problem with this one
X is actually identical to X. - the reasoning works because of the \subseteq sign, so you don't have to muddle with X being it's own subset and superset at the same time.

2) is not so obvious for me.
I understand that one consists of ordered pairs and the other of ordered triples. But I'm not sure if this affects anything.
if otherwise, then all ordered pairs would be ordered triples

3) is true because { 1 } is a subset of { -1 , 1}

4) is true because { -1 , 1 } is not a subset of { 1 }
technically - because -1 is a member of the {LHS} but not a member of {RHS}

x=-1 \in \left \{ (x,y): x^2 - x = 0 \right \} but x=-1 \notin \left \{ (x,y): x - 1 =0 \right \}

By The way: I was having trouble with 3 and 4. But I kinda got an insight while typing.
not sure if it could be more formal maybe.
... Sometimes explaining the problem leads to a solution :)

 

[Mark44]

This is more to see if I understand it or not.
There are four statements and I need to explain why they are true. (they all are)
I understand it why some of they are, but my answers just don't feel accurate/formal enough.

Homework Equations

1) \mathbb{R}^3 \subseteq \mathbb{R}^3
2) \mathbb{R}^2 \nsubseteq \mathbb{R}^3
3) \left \{ (x,y): x - 1= 0 \right \} \subseteq \left \{ (x,y): x^2 - x = 0 \right \}
4) \left \{ (x,y): x^2 - x = 0 \right \} \nsubseteq \left \{ (x,y): x - 1 =0 \right \}

The Attempt at a Solution

1) is true simply because X is a subset of X for any set X. no problem with this one

2) is not so obvious for me.
I understand that one consists of ordered pairs and the other of ordered triples. But I'm not sure if this affects anything.

3) is true because { 1 } is a subset of { -1 , 1}

Your reason is not especially relevant. {(x, y) : x - 1 = 0} is not a single number. This set and the other one in this problem are sets of ordered pairs. Think about what the graph of x = 1 looks like in the plane.

4) is true because { -1 , 1 } is not a subset of { 1 }

Your reason is not especially relevant.
As in #3, both sets are sets of ordered pairs.

By The way: I was having trouble with 3 and 4. But I kinda got an insight while typing.
not sure if it could be more formal maybe.

So the really troubling one is 2.

 

[U.Renko]

I see

so the correct answer for 3 would be

\left \{ ( 1, 0 ) \right \} \subseteq \left \{ (-1,0) , (1,0) \right \}

and for 4 it would be

\left \{ (-1,0) , (1,0) \right \}\nsubseteq \left \{ ( 1, 0 ) \right \}

thanks for helping!

 

[HallsofIvy]

No, those would not be the correct "answers" because your answer was supposed to be a reason. What you give in your last post is just an incorrect restatement of the problem.

(3) is true because every member of the set on the left (which is NOT just the pair (1, 0) but every pair of the form (1, y) where y can be any number) is also in the set on the right- which contains all pairs of the form (1, y) and (-1, y).

(4) is false because the set on the left contains some members of the form (-1, y) which arte not in the set on the right.

 

[U.Renko]

Oh I see.
basically what I was thinking about in my first post was that it simply represented the set of solutions to the equation X^2 - x = 0

But now that you mention I notice that in the Set builder notation, not only it mentions it is an ordered pair, but it puts no restraint on values of y.
So basically it can take any value of y...

Is that correct?
if so then I guess I might need to be more careful

 

[Mark44]

Yes, the set {(x, y) | x² - x = 0} represents two vertical lines in the plane. The equations of these lines are x = 1 and x = 0.