# Some basic questions on the way sets are defined

• sa1988
In summary: The Attempt at a SolutionQ.1I'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 elementsa) \emptyset\subset A - Trueb) \emptyset\in A - Truec) \{\emptyset\}\subset
sa1988

## 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.

sa1988 said:

## 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.

sa1988
PeroK said:
You shouldn't post two questions in 1.

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.

sa1988 said:
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.

sa1988
You could replace the elements of ##A## by some other names, solve the questions and replace them backwards.

sa1988 and PeroK
PeroK said:
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.

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

sa1988
fresh_42 said:
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## ..?

Last edited:
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
fresh_42 said:
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.

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.

Logical Dog and sa1988
Great, thanks a lot. All understood now.

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

fresh_42
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

sa1988 said:
f−1f−1

Last edited:

## What is a set?

A set is a collection of distinct objects or elements that are grouped together based on a common characteristic or property.

## How are sets defined?

Sets are defined by listing their elements within curly braces, separated by commas. For example, the set of even numbers can be defined as {2, 4, 6, 8, ...}.

## What is the cardinality of a set?

The cardinality of a set is the number of elements in the set. It can be represented by the symbol |A|, where A is the set. For example, if A = {1, 2, 3}, then |A| = 3.

## What is the empty set?

The empty set, or null set, is a set with no elements. It is represented by the symbol ∅ and is a subset of all other sets.

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

A set is a collection of elements, while a subset is a set that contains only some of the elements of another set. In other words, all elements of a subset are also elements of the original set, but not all elements of the original set are necessarily in the subset.

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