-1
$\begingroup$

Suppose $Y=\{y_1,\ldots,y_m\}$ partitions the set $X=\{x_1,\ldots,x_n\}$. I would like to define a function $y: X \to Y$ which returns $y \in Y$ if and only if $x \in y$. Is there a way to write this more formally?

I came up with

$$y(x) = \{y \in Y :y \ni x\}$$

or

$$y(x) = \{y:y \in Y \ \land \ x \in y \}$$

but I am not sure this makes sense.

$\endgroup$
3
  • 1
    $\begingroup$ In my opinion, it is a common misconception, that using words to describe something is informal. Defining $y: X \rightarrow Y$ as the function, that assigns $y(x)$ to the unique $y\in Y$ with $x\in y$, is already the most elegant way of describing the function. $\endgroup$ May 22 at 12:36
  • 2
    $\begingroup$ Also $\{y\in Y \: | \: y \ni x\}$ is not an element of $Y$, but rather a subset of $Y$. $\endgroup$ May 22 at 12:38
  • $\begingroup$ @LeanderTilstedKristensen, thanks for your help. You are perfectly right. $\endgroup$ May 22 at 13:08

2 Answers 2

1
$\begingroup$

The idea $$y(x) = \{y \in Y :y \ni x\}$$ is fine for me (but*). Since $Y$ is a partitions elements of it are not $\emptyset$ and for all $x\in X$ there are only one $y\in Y$ such that $x\in y$. If you like equivalence relation there is one connected with partition. So if $\sim\subset X\times X$ would be a equivalence relation such that $X/_\sim=Y$ your function could be represented as mapping $$X \ni x\mapsto [x]_{\sim}\in X/_\sim. $$

Edit*: Of course $\{y \in Y :y \ni x\}$ is a singleton of our point of interest as it was pointed out. I missed that sorry and thanks to @LeanderTilstedKristensen. However if you really want be fancy you can write $$y(x) = \bigcup\{y \in Y :y \ni x\}$$ in the spirit of descriptive set theory. But this is more like joke than real application thing.

$\endgroup$
4
  • $\begingroup$ As @LeanderTilstedKristensen pointed out, $ \{y \in Y :y \ni x\}$ is a subset of $Y$ rather than one of its elements, which makes it less formal than a simple word description. With resect to the equivalence relation, I just do not understand it, but it might be useful for someone else more advanced than me. $\endgroup$ May 22 at 13:11
  • $\begingroup$ @user_231578: With due respect to Marek, this answer is wrong for the reason pointed out by Leander. My opinion is that it would be better to not accept any of the answers, than to have an accepted answer that's in error. But it's your decision. $\endgroup$ May 22 at 13:26
  • $\begingroup$ @user_231578: and as for the equivalence relation part, why not try reading this. It's pretty basic, and it's a good idea to have in your tool belt. $\endgroup$ May 22 at 13:29
  • 1
    $\begingroup$ I edit some parts and make my mistake (overlook) clear now. $\endgroup$ May 22 at 13:30
1
$\begingroup$

(This answer just collects other responses into what I hope is a more coherent, single answer, with a few of my opinions added in.)

@LeanderTilstedKristensen 's second comment, "Also $\{y\in Y| x \in y \}$ is not an element of Y, but rather a subset of Y", is important - it means that your symbolic notations are actually incorrect: they specify a subset of $Y$, not an element of $Y$.

I also think that words are fine, and "$y(x)$ is the (unique) element of $Y$ that contains $x$" seems really clear, and anyone who has seen a partition before will understand it and know that it is a valid definition.

Of course, if a notation already exists we should use that instead of inventing something new. And if your audience is comfortable with the fact that partitions and equivalence relations are more-or-less the same thing, then @MarekKryspin 's suggestion is the best: write $[x]$ instead of $y(x)$.

If you might be using more than one partition, or really want to emphasize that the equivalence class derives from $Y$, the notation $[x]_Y$ would do the job. It's usually used as "$[x]_\sim$, where "$\sim$" is an equivalence relation, but again, partitions and equivalence relations are pretty interchangeable, so using $Y$ as the subscript would be clear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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