Let $m,n$ are give postive integer,and postive real numvers$x_{1}<x_{2}<\cdots<x_{m}$and give it ，
Define $A=\{x_{1},x_{2},\cdots,x_{m}\}$.
find the following minumum of the value
$$\sum_{k=1}^{n}\sum_{i=1}^{n}\sum_{j=1}^{n}\dfrac{\sigma{(A_{k}\bigcap A_{i}\bigcap A_{j})}}{\sigma{(A_{k}...
@user170039 If you want to ping somebody who was in this chatroom, it is only possible in this way for some period of time since the last message of that user. See also: Does ping work in chat with no autocompletion?
However, I do not think it is important, you probably do not care who answer your post. (I'd say that it is probably even better that I was not pinged.)
If you want to ping somebody who was not in the room for longer time, a possible way to go would be to reply to some of their messages. (Click on the small arrow next to a message in chat. And then click on "reply to this message".) I do not know about time limit for this type of notification.
@user170039 I do not see why you are using $z$ in the second formula. (It does not appear on the LHS.) I guess you wanted to write $$(x,y)\in G_f \Leftrightarrow (x\in X \land y=f(x)).$$
$x\times f(x)$ is a different notation for ordered pair $(x,f(x)$.
user131753
Yes. That we understand.
user131753
My friend's argument only is that the definition as written doesn't make any sense because roughly speaking for definition of a set we need to take an arbitrary element and then say explicitly what is the formula that this element satisfies. Munkres hasn't done that.
"Whenever a situation occurs where confusion is possible, we shall adopt a different notation for the ordered pair $(a,b)$, denoting it by the symbol $$a\times b$$ instead." (page 13 in 2nd Ed)
@user170039 Well, he has used set-builder notation.
@user170039 If you want to be very precise, the formula would be: $$(a,b)\in G_f \Leftrightarrow (\exists x\in X) (a,b)=(x,f(x)).$$
In general, if $g$ is some function and somebody defines a set $S=\{g(x); x\in X\}$, then you have $$s\in S \Leftrightarrow (\exists x\in X) s=g(x).$$
In this specific case the function would be $g\colon x\mapsto (x,f(x))$.
user131753
Exactly that's what it should be. Actually my friend argued that since Munkres stated the property of only the first component ($x\in X$) there is no reason to accept that the pair $(x_1,f(x_2))\not\in G_f$ even though $f(x_1)\ne f(x_2)$. Thanks anyway.
