Some basic questions on the way sets are defined

[sa1988]

Homework Statement

Homework Equations

The Attempt at a Solution

Q.1

I'm a little confused with how subsets and elements are defined in the case of the given set "A" as it seems to be a set of sets, so I'm going to throw my answers out and would appreciate any guidance on where I'm wrong (if anywhere).

A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\}

"A" has two elements

a) \emptyset\subset A - True
b) \emptyset\in A - True
c) \{\emptyset\}\subset A - Not true
d) \{\emptyset\}\in A - Not true
e) \{\emptyset, \{\emptyset\}\}\subset A - True
f) \{\emptyset, \{\emptyset\}\}\in A - True

Q.2

Not sure what's going on here either. I think the issue is in my own flawed understanding of the notation used in sets generally.

f : R \rightarrow R such that f(x) = x^{2}

My understanding thus far is that the cartesian product of two sets X and Y is:

X \times Y = \{(x,y) : x\in X, y\in Y\}

So in the case of f(x) = x^2, we have:

a) f((-1,2)) = (-1,2) \times (-1,2) = \big((-1,-1),(-1,2),(2,-1),(2,2)\big)

but then part of me wonders if I've got it all wrong and it should really just be f((-1,2)) = ((1,4)) ..??

And then for part b:

b) f((-1,2]) = ... ...

I don't really understand this at all since it has a square bracket which I'm led to believe means it represents a continuous interval of numbers not including that which is on the side of the curled bracket (according to this - interval notation). If that's the case, I don't know how to perform X \times Y in the way I defined above.

And then we have stuff to do with f^{-1} which is a whole other thing entirely.

(Just to check - am I right in saying that f^{-1} on a set Y is all the elements x \in X such that f(x) \in Y ??)

Or in other words: f^{-1}(Y) = \{ x \in X : f(x) \in Y \} - right?

Hints much appreciated, thanks.

 

[PeroK]

Homework Statement

Homework Equations

The Attempt at a Solution

Q.1

I'm a little confused with how subsets and elements are defined in the case of the given set "A" as it seems to be a set of sets, so I'm going to throw my answers out and would appreciate any guidance on where I'm wrong (if anywhere).

A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\}

"A" has two elements

a) \emptyset\subset A - True
b) \emptyset\in A - True
c) \{\emptyset\}\subset A - Not true
d) \{\emptyset\}\in A - Not true
e) \{\emptyset, \{\emptyset\}\}\subset A - True
f) \{\emptyset, \{\emptyset\}\}\in A - True

You shouldn't post two questions in 1.

I don't agree with your answers. Which ones are you sure about?

 

[sa1988]

You shouldn't post two questions in 1.

I don't agree with your answers. Which ones are you sure about?

To be honest I'm not sure on any of them so far.

My problem I think lies in that I don't understand the difference between \{\emptyset\} and \emptyset

To me it seems that one is the empty set, and the other is a set with the empty set in, which is just the empty set, so they're the same. This surely isn't the case because it makes things too trivial, but I don't know how else to interpret those two things.

Thanks.

 

[PeroK]

To be honest I'm not sure on any of them so far.

My problem I think lies in that I don't understand the difference between \{\emptyset\} and \emptyset

To me it seems that one is the empty set, and the other is a set with the empty set in, which is just the empty set, so they're the same. This surely isn't the case because it makes things too trivial, but I don't know how else to interpret those two things.

Thanks.

There is a difference between the element $a$ and the set containing $a$, denoted by $\lbrace a \rbrace$. Similarly for the empty set. The set containing the empty set is not empty! It has one member: the empty set.

 

[fresh_42]

You could replace the elements of $A$ by some other names, solve the questions and replace them backwards.

 

[sa1988]

There is a difference between the element $a$ and the set containing $a$, denoted by $\lbrace a \rbrace$. Similarly for the empty set. The set containing the empty set is not empty! It has one member: the empty set.

Excellent, thanks, I think it's all clicked now. Updated answers are below.

A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\}

"A" has two elements

a) \emptyset\subset A - NOT True because \emptyset is just an element, not a set, and therefore cannot be a subset
b) \emptyset\in A - True because it is an element of A
c) \{\emptyset\}\subset A - True since A = \Big\{\emptyset, \big\{\emptyset, \{\emptyset\}\big\}\Big\} = \{\emptyset\}\cup \Big\{\big\{\emptyset,\{\emptyset\}\big\}\Big\}, so clearly \{\emptyset\}\subset A
d) \{\emptyset\}\in A - NOT true, this set is not en element of A.
e) \{\emptyset, \{\emptyset\}\}\subset A - NOT True, it is only an element.
f) \{\emptyset, \{\emptyset\}\}\in A - True.

 

[fresh_42]

To a): Can you tell all subsets of $A=\{1,2\}\, ?$ How many are there? Why doesn't the argument under c) apply?

 

[sa1988]

To a): Can you tell all subsets of $A=\{1,2\}\, ?$ How many are there? Why doesn't the argument under c) apply?

Hmm, I hope I'm right in saying \emptyset, \{1\}, \{2\}, \{1,2\} are subsets of \{1,2\}.

This now leads me to question my result for part a) however...

Another way I'm seeing it is that the empty set is {} , hence {}, {1}, {2} and {1, 2} are subsets of {1,2}

So for { 0, {0,{0}} }, the subsets are {}, {0}, { {0,{0}} } and { 0, {0,{0}} }

and {} is the empty set $\emptyset$

hence $\emptyset \subset A$ ..?

 

[fresh_42]

Yes, the empty set is always a subset, regardless what is in the set. You said it already under c) : $A=A\cup\{\} \Rightarrow \{\} \subseteq A$.

 

[sa1988]

Yes, the empty set is always a subset, regardless what is in the set. You said it already under c) : $A=A\cup\{\} \Rightarrow \{\} \subseteq A$.

Great stuff, thanks.

So just to check - the answers I gave for a) to f) were correct apart from part a), which should have been 'true' since $\emptyset$ is always a subset of any set.

 

[fresh_42]

If I didn't become confused myself, yes. In the original post c) and e) were wrong. e) could be "repaired" to be true with an additional bracket around: $\{\{\emptyset, \{\emptyset\}\}\} \subsetneq A\,,$ because the element would be then in a set.

 

[sa1988]

Great, thanks a lot. All understood now.

I'll have another look at Q2 tomorrow, probably in a new thread as requested.

 

[Logical Dog]

I think, for d, at least for F(x) is equivalent to

\left (-\infty ,0 \right ]\times [0, \infty) but don't know how to find f-1, hinki it is an emppty set

f−1f−1