# Understanding some basic set theory stuff.

• U.Renko
The set {(x, y) | x - 1 = 0} is the single point (1, 0). The point (1, 0) is on both lines, so the set on the left is a subset of the set on the right.
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.

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 :)

Last edited:
U.Renko said:
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.
U.Renko said:
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.
U.Renko said:
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.

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!

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.

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

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.

## 1. What is set theory?

Set theory is a branch of mathematics that deals with the study of sets, which are collections of objects or elements. It provides a foundation for many areas of mathematics and has applications in fields such as computer science and statistics.

## 2. Why is understanding set theory important?

Set theory is important because it helps us understand the relationships between different sets and how they can be manipulated using mathematical operations. It also provides a framework for defining and analyzing mathematical concepts and can be applied to various real-world problems.

## 3. What are the basic elements of set theory?

The basic elements of set theory are sets, elements, and operations. Sets are collections of objects or elements, and elements are the individual objects within a set. Operations, such as union, intersection, and complement, are used to manipulate sets and create new sets.

## 4. What is the difference between a set and a subset?

A set is a collection of objects, while a subset is a set that contains only elements that are also included in another set. In other words, a set is a larger collection that may contain multiple subsets within it.

## 5. How can set theory be applied in real life?

Set theory has various applications in real life, such as in computer science for database management and data analysis. It is also used in statistics for probability and data analysis. In addition, set theory can be applied in decision-making processes, such as creating Venn diagrams to compare different options.

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