Given the function $f(x) = \sqrt x$
and given that $S_1 =(-2, 2)$ and $S_2=( -1 , 3)$
Are these two equivalent?
$f(S_1∩S_2)$ and $f(S_1)∩f(S_2)$?
I was wondering because generally it's false, but in this case it seems to be true.
Given the function $f(x) = \sqrt x$
and given that $S_1 =(-2, 2)$ and $S_2=( -1 , 3)$
Are these two equivalent?
$f(S_1∩S_2)$ and $f(S_1)∩f(S_2)$?
I was wondering because generally it's false, but in this case it seems to be true.
In what I've written below, I've assumed you mean $f$ to be the real square root function $(0,\infty)\rightarrow \mathbb{R}$, and that by $f(X)$ you mean the image under $f$ of all elements of $X$ in the domain of $f$. You should be aware that the usual convention is that $f$ can only be applied to subsets of its domain, so $f(S_1)$ is not defined, since it contains negative numbers.
If this is not what you mean, i.e. if you intend $f$ to be a complex square root function, please let me know.
Yes, in this case, $f(S_1\cap S_2) = f(S_1)\cap f(S_2)$. You point out that in general, $f(X\cap Y)$ may not be equal to $f(X)\cap f(Y)$. This is also true, but let's explore it.
Suppose $a\in f(X\cap Y)$. Then there is some element $b\in X\cap Y$ such that $f(b) = a$.
Now $b\in X$, so $a = f(b) \in f(X)$, and $b\in Y$, so $a = f(b) \in f(Y)$.
So $a\in f(X) \cap f(Y)$. This shows that $f(X\cap Y) \subseteq f(X) \cap f(Y)$.
What could go wrong in trying to prove the converse: that $f(X) \cap f(Y) \subseteq f(X\cap Y)$?
Well, if $a\in f(X) \cap f(Y)$, then $a\in f(X)$, so there is some $b\in X$ with $f(b) = a$, and $a\in f(Y)$, so there is some $b'\in Y$ with $f(b') = a$.
But the elements $b$ and $b'$ could be different, i.e. $b$ may not be in $Y$ and $b'$ may not be in $X$. There may be no element in $X\cap Y$ which maps to $a$.
But the only way this could happen is if there are two distinct elements $b$ and $b'$ which both map to $a$. If there is only one element which maps to $a$, then it must be in both $X$ and $Y$.
In conclusion, if $f$ is injective, then $f(X\cap Y) = f(X) \cap f(Y)$. Since the square root function is strictly increasing, it is injective, and hence the equality holds not just for the sets you give as examples, but all $S_1$ and $S_2$ subsets of $\mathbb{R}$.