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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.

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Are you sure you want to be drawing a square root from negative numbers? – joriki Oct 26 '12 at 1:44
i know it's not defined, but i don't think it's necessarily wrong. – Gladstone Asder Oct 26 '12 at 1:57
up vote 0 down vote accepted

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 $\mathbb{R}$.

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From the comments, OP wants to take square roots of negative numbers. At that point, the part of the argument that relies on the square root function being strictly increasing falls down. – Gerry Myerson Oct 26 '12 at 3:14
His comment "I know it's not defined, but I don't think it's necessarily wrong" suggests to me the interpretation I've given in my first sentence - that is, that $f$ is a function $(0,\infty)\rightarrow \mathbb{R}$. – Alex Kruckman Oct 26 '12 at 3:19
You may be right. Perhaps OP will clarify just what's intended for negative arguments. – Gerry Myerson Oct 26 '12 at 3:39

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