How to write this set in set-builder notation?

[Math100]

Homework Statement:: Write each of the following sets in set-builder notation.
Relevant Equations:: None.

{2, 4, 8, 16, 32, 64...}
=2·{1, 2, 4, 8, 16, 32...}
What should be the next step in this work?

 

[fresh_42]

I assume by set-builder notation you mean ##\{x \in S\,|\,P(x)\}##, the set of all ##x## from ##S## which satisfy the predicate ##P(x)##. E.g. the set of all primes is ##\{2,3,5,7,11,13,17,19,\ldots\}=\{x\in \mathbb{Z}\,|\,x>0 \text{ and } x \text{ is prime }\}.##

This is a guess, but if so, what would ##P(x)## be in your example?

 

[Math100]

I assume by set-builder notation you mean ##\{x \in S\,|\,P(x)\}##, the set of all ##x## from ##S## which satisfy the predicate ##P(x)##. E.g. the set of all primes is ##\{2,3,5,7,11,13,17,19,\ldots\}=\{x\in \mathbb{Z}\,|\,x>0 \text{ and } x \text{ is prime }\}.##

This is a guess, but if so, what would ##P(x)## be in your example?

The P(x) would be 2^x. But how should I write a proof that comes to this solution involving this predicate P(x)?

 

[fresh_42]

The P(x) would be 2^x. But how should I write a proof that comes to this solution involving this predicate P(x)?

The notation usually depends on the author. There is no single right version. You could write is as
\begin{align*}
S_1&=\{2,4,8,16,32,\ldots\}\\
S_2&=\{2^n\,|\,n\in \mathbb{N}\}\\
S_3&=\{x\in \mathbb{Z}\,|\,\exists \,n\in \mathbb{N}\, : \,x=2^n\}\\
S_4&=\{x\in \mathbb{Z}\,|\,\log_2(x)\in \mathbb{N}\}
\end{align*}

The first one is the one in the question.
The second one is the one I would use.
The third one is the most formal one. ##\exists## reads: exists or there is and : means such that.
The fourth one is one I chose to have another example.

 

[Math100]

The notation usually depends on the author. There is no single right version. You could write is as
\begin{align*}
S_1&=\{2,4,8,16,32,\ldots\}\\
S_2&=\{2^n\,|\,n\in \mathbb{N}\}\\
S_3&=\{x\in \mathbb{Z}\,|\,\exists \,n\in \mathbb{N}\, : \,x=2^n\}\\
S_4&=\{x\in \mathbb{Z}\,|\,\log_2(x)\in \mathbb{N}\}
\end{align*}

The first one is the one in the question.
The second one is the one I would use.
The third one is the most formal one. ##\exists## reads: exists or there is and : means such that.
The fourth one is one I chose to have another example.

So when solving this type of problem, is there a need to show work by writing a proof? Or just write out the answer by figuring out what the predicate P(x) is?

 

[fresh_42]

So when solving this type of problem, is there a need to show work by writing a proof? Or just write out the answer by figuring out what the predicate P(x) is?

Yes, the latter. Figuring out what ##P(x)## is and then writing it in the required style. But all this assumes that my interpretation of set-builder notation is correct.

It actually makes sense because ##S=\{x\in \mathbb{Z}\,|\,\exists\;n\in \mathbb{N}\, : \,x=2^n\} ## is exact, i.e. there is only one possible set. If we write ##\{2,4,8,16,32,\ldots\}## then this is ambiguous. It looks like the set ##\{2,4,8,16,32,64,128,256,\ldots,2^n,\ldots\}## but it could as well go on completely differently, e.g. ##\{2,4,8,16,32,33,34,35,36,\ldots\}##. A finite number of elements does not determine an infinite sequence. The formal notation, however, says how all elements look like and infinity is hidden in ##\mathbb{N}##, not in dots.

 

[Math100]

So how to write this proof then? How should I start?

 

[fresh_42]

I don't think that it is more than simply saying
$$
A:=\{2,4,8,16,32,\ldots\} = \{x\in \mathbb{Z}\,|\,\exists n\in \mathbb{N}\, : \,x=2^n\}=:B
$$

Sure, you can pump it up, but that's a bit too much in my opinion.

If you want to prove that two sets are equal, say ##A=B##, then the usual way is to show ##A\subseteq B## and ##B\subseteq A##. For e.g. ##A\subseteq B## we choose an arbitrary element ##a\in A## and show that it is an element of ##B##, that it satisfies the predicate of ##B##.

##A\subseteq B\, :##

Let ##a\in A.## ...
(The important point here is that it must be arbitrary, fixed, but any element of ##A##. We are not allowed to impose any restrictions, since we want to have all elements of ##A##.)
... Then ##a## is a power of ##2## because every element of ##A## is twice its predecessor, starting with ##2##. Hence there is a natural number ##n\in \mathbb{N}## such that ##a=2^n##, i.e. ##a## satisfies the defining predicate of ##B.##

##B\subseteq A\, :##

Let ##b\in B.## Then ##P(b)## is true, i.e. ##b=2^n## for some ##n\in \mathbb{N}.## But the powers of ##2## are exactly the elements of ##A##, hence ##b\in A.##

From ##A\subseteq B\subseteq A## follows ##A=B.##

This would have been formal proof, but it is not really necessary in such a simple case. However, there are cases, where it is not obvious that ##A=B##.

 

[Math100]

Thank you.

 

[Mark44]

Write each of the following sets in set-builder notation.

So how to write this proof then? How should I start?

You're not asked to write a proof of anything, but rather to write the given set in set-builder notation. If the problem included wording such as "Prove that ... " or "Show that ...", then some sort of proof would be required.

I don't think that it is more than simply saying
$$
A:=\{2,4,8,16,32,\ldots\} = \{x\in \mathbb{Z}\,|\,\exists n\in \mathbb{N}\, : \,x=2^n\}=:B
$$

I agree.

 

[Math100]

You're not asked to write a proof of anything, but rather to write the given set in set-builder notation. If the problem included wording such as "Prove that ... " or "Show that ...", then some sort of proof would be required.

I agree.

Thank you.