2
$\begingroup$

I recently need to write some proofs involving the relationship between donmain, co-domain and range, which I am little confused about, for example:


1.If $f:A\rightarrow{R},g:B\rightarrow{R},g \circ f:C\rightarrow{R}$

Then we have $C = range(f)\cap A \cap B$

2.If we have some $c\in C$, then also $f(c)\in B$


My questions:

1.Is these statements true, if so, is it possible to prove it?

Or if it's false, what would be the counter example and the right relationship?

2.Any other result we can conclude from the condition?

Any help would be appreciated.


Definitions I'm using:

$f:A\rightarrow{B}$:

$f$:domain $\rightarrow$ co-domain

domain:

Subset of R that f is defined on

(for example, domain of $\frac{1}{x}$ is R without $0$)

co-domain:

R as default

range:

Outputs of f as a subset in co-domain

$\endgroup$
0
1
$\begingroup$

The requirement here is that we need $g \circ f$ to make sense and that is possible when $f(C) \subseteq \text{domain}\ (g) = B$. So if $C = \text{range}\ (f) \cap A \cap B,$ then $x \in C \Rightarrow x \in A \Rightarrow f(x) \in \text{range}\ (f)$. This is the best we can conclude from here. So there may very well be some elements in range$\ (f)$ such that they are not in domain $(g)$. Hence we can not define $g \circ f$ properly from this.

Consider for example: $f : \mathbb{R} /\{0\} \longrightarrow \mathbb{R}$ defined as $f(x) = \frac{1}{x}$ and $g: \mathbb{R}/\{1,2\} \longrightarrow \mathbb{R}, g(x) = \frac{1}{x^2-3x+2}$. Then, we can see that $g \circ f: \mathbb{R}/\{\frac{1}{2},1\}\longrightarrow \mathbb{R},g(f(x)) = \frac{x^2}{1-3x+2x^2} = \frac{x^2}{(2x-1)(x-1)} $. For the moment we just focus on the domains.

We can here see that $C = \mathbb{R}/\{1, \frac{1}{2}\} \ne \text{range}\ (f) \cap A \cap B$ as $\frac{1}{2} \in \text{range}\ (f)\cap A\cap B $ but doesn't in $C$.

We can construct many examples like this. The key is to understand when is $g \circ f$ defined. Hope this helps.

$\endgroup$
4
  • $\begingroup$ Yes, this helps!! would it be right if we rewrite it as $C = range(f)\cap A \cap B$? $\endgroup$ – Manx Jun 12 '19 at 9:44
  • $\begingroup$ @Subcat I have edited the answer as per the question, I guess you came up with this whence I gave the first example where $C = \text{range} \ (f) \cap A \cap B$. But that is not the point, I think you should understand how the composition works. Then you can form many examples as the one mentioned above in the answer. :) $\endgroup$ – Rick Jun 12 '19 at 14:45
  • $\begingroup$ Also please correct the spelling of domain in the question and the title. $\endgroup$ – Rick Jun 12 '19 at 14:48
  • $\begingroup$ I think you are right , but $range(\frac{1}{x}) = R/\{0\}$ $\endgroup$ – Manx Jun 14 '19 at 0:02

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.