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.