Timeline for $f[A\cup{B}]=f[A]\cup{f[B]}$ [duplicate]
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2014 at 18:12 | history | closed | Nick Peterson Brian Rushton Olivier Bégassat Dan Rust AlexR | Duplicate of Prove $f(S \cup T) = f(S) \cup f(T)$ | |
| Jan 5, 2014 at 17:41 | review | Close votes | |||
| Jan 5, 2014 at 18:12 | |||||
| Dec 10, 2013 at 15:20 | history | edited | amWhy | CC BY-SA 3.0 | formatting |
| Sep 17, 2013 at 15:08 | vote | accept | Iceman | ||
| Sep 17, 2013 at 15:03 | history | edited | Iceman | CC BY-SA 3.0 | deleted 27 characters in body |
| Sep 17, 2013 at 15:03 | comment | added | anon | Consider the act of putting stickers on apples that are already in a bag. After you're done, the set of apples will be comprised of one stickered apple for each apple contained in the original bag. | |
| Sep 17, 2013 at 15:03 | comment | added | Iceman | In hindsight I thought it was needed but I see what @ThomasAndrews is saying in that, mentioning that i'm letting $x\in{A\cup{B}}$ implies that I'll be using a particular $x$ for a proof...I will remove that. Thanks everyone! | |
| Sep 17, 2013 at 14:59 | comment | added | D.L. | I was speaking of the sentence after the point. I don't understand the utility of the "let $x\in A\cup B$". | |
| Sep 17, 2013 at 14:57 | comment | added | mdp | If you're worried about how $f$ interacts with the set builder notation, you can avoid it completely by proving $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A)\cup f(B)\subseteq f(A\cup B)$. But as has already been said, your existing argument is correct. | |
| Sep 17, 2013 at 14:56 | answer | added | amWhy | timeline score: 9 | |
| Sep 17, 2013 at 14:55 | comment | added | Thomas Andrews | You sort of misuse $x$ in that first line. "Let $x\in A\cup B$" means "Let $x$ be a particular element of $A\cup B$." Then the next sentence you write, $f[A\cup B]=f\{x\mid x\in A\text{ or } x\in B\}$, which actually is a different usage of $x$. | |
| Sep 17, 2013 at 14:52 | comment | added | D.L. | Indeed, it's not the case : your second line is completely true. | |
| Sep 17, 2013 at 14:49 | history | asked | Iceman | CC BY-SA 3.0 |