Question

Problem: Prove that, if f and g are functions, then f∩g is a function by showing that f∩g=g|A where A={x:g(x)= f(x)}.
I am uncertain how to tackle this. Do I start by proving f(x)=g(x) or f∩g? Or something else entirely? Any assistance would be appreciated.

Answer 1

( ͡° ͜ʖ ͡°)

Answer 2

@AlexandervonHumboldt2

Answer 3

He should be able to help. :P

Answer 4

Thank you.

Answer 5

No problem.

Answer 6

Thanks Alex :P

Answer 7

f and g are functions. Intersection of them is where f(x)=g(x). The point where the graphs intersect is their intersection. Now we have that f(x)=g(x), A={x|g(x)=f(x)}. Thus \[f∩g=g|A\]

i'm not sure if i provided enough strict proof.

let me have someone to check

Answer 8

@Michele_Laino @freckles can you guys check me?

Answer 9

could it also be also that \[f \cap g = f|A\]

Answer 10

yeah @pgpilot326 is correct

Answer 11

since f and g are functions, each input has only 1 output so with A = {x: g(x) = f(x)} each point in A is unique and has only a single output

Answer 12

i should say that each point in x is unique and when put into either f or g leads to a single output since both f and g are functions

Answer 13

question seems weird to me
it seems like
\[A \in \mathbb{R} \text{ but } f,g \neq \mathbb{R} \\ \text{ like } f,g \text{ could be in } \mathbb{R} \times \mathbb{R}\]

Answer 14

unless g|A doesn't mean what I think it means

Answer 15

It couldn't mean what I think it means because that makes the question not make sense :p
you know the whole A|B={x:x is in A but not in B thing}

Answer 16

if f and g are in 2D, then so is x

Answer 17

Thank you for the assistance @AlexandervonHumboldt2 @pgpilot326

So, when writing out the proof, I could start by showing that, since f(x)=g(x), then for any x that is an element of A, x is an element of both f(x) and g(x). Then from there I could show that, from that information, that the restriction of g to A is where x is an element of both f and g. After that, I could then show that g|a is equal to the intersection of f and g then. Would that work?

Proofs as a whole are not my strong suit and this one is really stumping me, so I apologize in advance if I messed any reasoning up there or for my lack of understanding.

@freckles For the problem, g|a is the restriction of g to A. Sorry, I should have clarified that when writing out the problem the first time!

Answer 18

x would just be a vector and f(x), g(x) would be a map to the same dimension.

you can't assume that f = g... only that if there are points for which f = g, then x is in A

Answer 19

Ah, okay.

Answer 20

A must contain all of the points (x) for which f = g. that is how it is defined, only you have to define it that way in your proof (since it's not a given).

Answer 21

ah gotcha
So I'm going to write that set a little differently\[g|A=\{(x,g(x)|f(x)=g(x)\} \]


\[\\ \text{ Let } (p,q) \in f \cap g \\ \text{ then } (p,q) \in f \text{ and } (p,q) \in g\]
then you can say .. something about the set g|A


So what I think you should do is... it show both ways...
I mean
show
\[g|A \subset f \cap g \\ \text{ and show } f \cap g \subset g|A \\ \text{ then showing both of those show } g|A=f \cap g\]

Answer 22

I will finish the first way I was talking about...
\[f(x)=g(x) \text{ when } x=p \\ \text{ and } g(p)=q \text{ since } (p,q) \in g \\ \text{ so } (p,q) \in g|A=\{(x,g(x))|f(x)=g(x)\} \\ \text{ and so } f \cap g \subset g|A\]

Answer 23

then you do the other way

Answer 24

@freckles Okay, that makes it a lot clearer! I see how that works now and will work out g|Af⊂∩g then.

Thank you all again for your help with this problem. I really appreciate it.

Answer 25

np