functions stuff

[silk cedar]

If we have the function x-->2x, with domain [0,1] and codomain [-1,7), does it change anything about the function? It still should be bijective, right? The codomain might be [-1,7), but the range is still [0,2], or am I thinking it wrong?

[ruby rampart]

I believe you are right

Like you are right
A function is basically just a set of pairs where every input has just one output
So if you make your codomain bigger then your range it doesn’t change the actual function

[silk cedar]

hmm okay okay

so a function being bijective or not depends on whether the codomain is the range, and if the function is 1-1, right?

[ruby rampart]

Yeah sure

The codomain is the range is kind of a mid way to think about it
Though it’s true
I guess if it helps for your understanding

Just because for proofs you use
for every y in the codomain there exists x in the domain such that f(x)=y
To prove surjectivity

Like that’s what needs to come to your mind first

And for injectivity the common proof method is just
For any a, b in your domain
If f(a)=f(b), then a=b

[ruby rampart]

Though I think it’s more common to present injectivity for the definition as the contrapositive in proofwriting books

[ivory pecan]

I've heard range used to describe image too, so I would be careful

Easiest to just say domain, codomain

[ivory pecan]

It is true that $[0,1]$ and $[1,7)$ have the same cardinality (there \emph{is} some bijection between them) but this function is not surjective on $[1,7)$. For example, what is the preimage of 6?

[deep hatch]

SelahW

[ivory pecan]

Any interval of the reals has cardinality continuum so this statement is not that interesting

[silk cedar]

I agree. Our prof told us that too

[silk cedar]

what do you mean by preimage?

btw srry i slept for like 11 hrs yesterday lol

I would guess that for our function, a preimage for 6 would not exist

because there exists a preimage only for every element of the actual image of the function's domain

[ivory pecan]

For $y\in Y$, the preimage under $f:X\to Y$ is the set ${x\in X\mid f(x)=y}$.

[deep hatch]

SelahW

[ivory pecan]

Right, this is the definition

It gives us some information though. If each preimage is a singleton, the function is injective. If the preimage of every element is nonempty, the function is surjective

[silk cedar]

so the preimage is kind of the inverse image of the function, right? Does it have any extra properties? I realized that the inverse image of a function f: A-->B should be C(subset of B)-->A and |C|\leq |A|

,tex
the inverse image of $ f: A \mapsto B $ is $ f^{-1}: C\mapsto A $ where $ C\subseteq B $

[deep hatch]

fijokazż

[silk cedar]

what do you mean by "each preimage" here?

i think I get what you mean about the properties, but what if we don't know anything about the original function?

I had an exercise that hinted that if an inverse image exists, then it has some extra property compared to a function that we dont know whether it is injective, or surjective

[silk cedar]

sorry again for replying so out of time, I get busy at weekdays and forget about my questions posted here

[silk cedar]

.close