Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For example:

$$a = \{\mathbf{true, true, false}\}$$ $$b = \{ 0.8, 0.8, 0\}$$

In this case each $\bf true$ in $a$ has become a $0.8$ in $b$ wheras the $\bf false$ values have resulted in $0$ s.

What would I write to express this for creating $b$ from any set $a$ as in the above example?

share|improve this question
    
What do you mean by creating? I suppose you could write $b=f[a]$ where $f(true)=0.8,f(false)=0$. –  tomasz Aug 21 '12 at 14:40
    
Create like $a = 4$ or $a = b$ if there was no $a$ before that definition. What do the [] represent? –  alan2here Aug 21 '12 at 14:46
    
What's wrong with $b = \{ 0.8, 0.8, 0\}$? $[]$ represent image. See e.g. math.stackexchange.com/questions/109942/…. –  tomasz Aug 21 '12 at 14:51
    
the '$b =$' line should work with any $a$, not just this example $true, true, false$. –  alan2here Aug 21 '12 at 14:59
    
Then you should edit your question to make it clear what you want exactly. If it's image, then my comment already answers your question (and I will post it as an answer so that you can accept it). –  tomasz Aug 21 '12 at 15:00
show 1 more comment

2 Answers

up vote 3 down vote accepted

First observe that sets ignore repetitions. Namely, $\{x,x,y\}=\{x,y\}$.

Now you have some $f\colon a\to b$, and you want to say that $f(x)=0.8$, then you can write $f^{-1}(0.8)=x$, or if you prefer $x\in f^{-1}[0.8]$ to suggest that $f(x)=0.8$ but possibly other elements are mapped to $0.8$ as well.

In the same manner, $y\in f^{-1}[0]$ is to say that $f(y)=0$.

To put this into your particular example, $\text{true}=f^{-1}(0.8),\text{false}=f^{-1}(0)$.

share|improve this answer
    
Sorry if this is obvious, is it also correct as f(0.8) = true, f(0) = false? Perhaps $a$ is a list and not a set because its entites are ordered and it contains duplicates? –  alan2here Aug 21 '12 at 15:09
    
@alan2here: If those are lists, then the function from the underlying set can be extended to the lists in the obvious way. –  Asaf Karagila Aug 21 '12 at 15:10
    
Given a = {true, true, false}, is the following ok to change the values in $a$? $a'(0.8) = true$, $a'(0) = false$. Also, I'd prefer to avoid the part in your example that appears as if $f$ to the power of -1 is being calculated. –  alan2here Aug 21 '12 at 15:12
    
@alan2here: I'm not sure what that means. However if you wish that all the elements which have the value $0.8$ in the first list will correspond to elements with value $\text{true}$ in the second list, then setting $f(0.8)=\text{true}$ is fine. –  Asaf Karagila Aug 21 '12 at 15:14
    
Thank you. $a'$ means "$a$ afterwards" or "$a$ becomes". –  alan2here Aug 21 '12 at 15:17
add comment

First, note that you are working with multisets rather than sets, since you are allowing repeated elements.

What you have in the example is a function $f: a \rightarrow b$ defined by $$ f(x) = \begin{cases} 0.8 & \text{if } x = \text{true}\\ 0 & \text{if } x = \text{false}. \end{cases} $$

You can replace the domain $a$ with any multiset of $0.8$'s and $0$'s. If you want to introduce new elements, you will need to specify how $f$ acts on them.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.