Question

Let X be a set and ∼ a relation on X. Define
N = {x ∈ X : ¬(x ∼ x) }.
Let
B = {b ∈ X : (∀n ∈ N)(b ∼ n) ∧ (∀n ∉ N)[¬(b ∼ n)] }.
Show that B = ∅

Answer 1

does that mean N is a set of all elements that don't have a reflexive mapping in the relation ?

Answer 2

from the definition of the set, yes

Answer 3

\(\rm B = \{b \in X : \bbox[yellow]{(\forall n \in N)(b \sim n)} \land (\forall n \not \in N)[\neg (b \sim n)] \}\)

B is a set of elements which satisfy below two properties :
1) there is a mapping from an element in B to every element in N (highlighted)
2)

Answer 4

@ChristopherToni i give up

Answer 5

yes, but i'm having difficulty saying it in logical form. I'm trying to prove by contradiction. But the definition of B is confusing.

I start off with B nonempty. b in B means b has to satisfy (∀n ∈ N)(b ∼ n) ∧ (∀n ∉ N)[¬(b ∼ n)]. I'm stuck on the next step

Answer 6

:O what the???

Answer 7

ASCII symbols aren't apparently recognized by the LaTeX system >_>

Answer 8

yeah B is very confusing.

use latex to avoid those diamonds,
copy paste the code from my reply

Answer 9

but anyway, b has to satisfy both statements in set B

Answer 10

The idea is to connect N and B somehow

Answer 11

Both of them are sets. So I have to come up with a logical statement that connects N and B

Answer 12

B and N are disjoint because if there is an element common to both, then it needs to be reflexive which is not possible in N

Answer 13

so the possible elements in B are from : X-N

Answer 14

@rsadhvika I can't see the connect between N and B base on your argument. I feel like there are many holes to fill in first

Answer 15

suppose \(a \in (B \cap N)\)
that means \((a\sim a)\) by definition of \(B\) (first statement)
the same is impossible according to definition of set \(N\)

Answer 16

@rsadhvika but that only shows a \in (B \cap N) is empty. Not B is empty right?

If I let B = {1,2,3} and N = {4,5,6}

a \in (B \cap N) is certainly empty but N is not empty

Answer 17

I can't even type latex here XD

Answer 18

thats right, all i said earlier was possible elements in B are from : X-N

N is out.

Answer 19

never said B is empty, did i ?

Answer 20

we can use second statement to conclude B is empty
lets see

Answer 21

This is how I see things:

Let \(R\) be this relation (as a set). Take \(N = \{x\in X: (x,x)\notin R\}\).

Now, if \(b\in X\) so that \(\forall\,n\in N,\,(b,n)\in R\), then \(b\neq n\implies b\notin N \).

However, if \(b\in X\) so that \(\forall\, n\notin N,\,(b,n)\notin R\), then \(b=n\implies b\in N\). Thus, \(B=\{b\in X: b\in N \wedge b\notin N\} = \emptyset \).

Please point out to me if you think I've made a faulty assumption or something (I have a headache right now, so it's possible).

Answer 22

@ChristopherToni I think you're on something here. But let me make sense of your argument first. I believe you're trying to show that b \neq n in the first condition of B?

Answer 23

I was trying to show that the first condition of set B's definition implies that \(b\notin N\).

Answer 24

Just really quick, how did you type latex without using *Sigma* Equation? I tried but it didn't work

Answer 25

wrap them with between :

```
\(


\)
```

Answer 26

or \ [ \] (with no space for displayed equations)

Answer 27

let \[ x\in N\]

Answer 28

what are there two lines???

Answer 29

why*

Answer 30

displayed inline would be something like \ (\dislaystyle <LaTeX goes here> \)

Answer 31

use \( \) for inline latex

Answer 32

\displaystyle *

Answer 33

use `\( \)` for inline latex

Answer 34

how do you do that code box? XD

Answer 35

anythign between backticks gives that graybox..

Answer 36

copy paste exact below :

```
`this` will have a gray background
```

Answer 37

ok, let me try a gain.

let \( x \in N\)

Yay, it works! ok back to the problem lol

Answer 38

congratulations ! ok back to problem i think i understood chris's master piece, it looks good to me xD

Answer 39

@ChristopherToni how did you conclude \(b \neq n\) for the first condition??

Answer 40

Recall that we defined \(N\) to be the values of \(x\in X\) such that \((x,x)\notin R\). If \(b=n\), then by definition of \(N\), we would see that \((b,n)\notin R\implies b\in N\). However, since we instead have \((b,n)\in R\), then \(b\neq n\); hence we have that \(b\notin N\).

I'm having a hard time explaining things with this headache of mine, so I hope this is making sense.

Answer 41

Whoah!!! I see it now!! thank you show much :DDD

Answer 42

No problem! :-)