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I only realised that equality may fail in $f[A]\cap f[B]\supseteq f[A\cap B]$ (i.e., that we can have $A,B,f$ for which $f[A]\cap f[B]\neq f[A\cap B]$) after checking the answer.

I don't see any problems in this spurious string of equivalences which seems to imply that $f[A]\cap f[B]$ is always equal to $f[A\cap B]$. So where does it fail?

I'm not enquiring about counterexamples for the failure of $f[A]\cap f[B]\subseteq f[A\cap B]$.

I'm also aware of 9.29: If $f$ is a one-to-one function, then $f[A]\cap f[B] = f[A\cap B]$. Where can this be applied to fix the broken string? Please advise me of other problems.

$\begin{align} \text{ A function } f(x)\in f(A\cap B) &\iff x\in A\cap B\\ &\iff x\in A \; \wedge \; x\in B\\ &\iff f(x)\in f(A) \; \wedge \; f(x)\in f(B)\\ &\iff f(x)\in f(A) \cap f(B) \end{align}$

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It is always the case that $f[A\cap B]\subseteq f[A]\cap f[B]$. If $y\in f[A\cap B]$ then there is an $x\in A\cap B$ such that $f(x)=y$. It follows that $f(x)\in f[A]$ and $f(x)\in f[B]$.

However, it is not always the case that $f[A\cap B]\supseteq f[A]\cap f[B]$. If we look at your purported proof, the error is in the first inference (working from the bottom to the top): Suppose that $y \in f[A]\cap f[B]$. Then there exist $x_1\in A$ and $x_2\in B$ such that $f(x_1)=f(x_2)=y$. However, unless $f$ is injective, you cannot assume that $x_1=x_2$.


There's another error too: $f(x)\in f[A\cap B]$ is not equivalent to requiring that $x\in A\cap B$. Again, this condition only holds for all $x$ in the case that $f$ is an injection.

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  • $\begingroup$ You wrote that equality can fail in $$ f [A]\cap f [B] \subseteq f [A\cap B] $$ but that doesn't seem to be what you meant. Saying equality can fail means that the inequality always hold but is sometimes strict. That is not the case in this example. The reverse inequality is the one which always holds. The fact that the inequality you quoted does not always hold is proof that the reverse inequality is sometimes strict. $\endgroup$ – Unwisdom Apr 12 '14 at 11:58
  • $\begingroup$ You assumed that $ x_1=x_2$ when you gave them both the same name, $ x $. The fact that $ x_1\in A $ and $ x_2\in B $ does not tell you that $ x\in A\cap B $ since there is no term $ x $, unless you are assuming that $ x $ is a name for *both* $ x_1$ and $ x_2$. $\endgroup$ – Unwisdom Apr 12 '14 at 12:03
  • $\begingroup$ +1. Thank you very much. Please advise if you mind the recent edit to elucidate what's 'misstated' and my attempt. $\endgroup$ – Greek - Area 51 Proposal Apr 13 '14 at 10:26
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Let $$f:x\mapsto x^2$$ and $$A=(0,\infty)\quad;\quad B=(-\infty,0)$$ then we have $$f(A\cap B)=\emptyset\subset f(A)\cap f(B)=(0,\infty)$$

Edit In your proof you have two mistakes:

  • The first by writing $$f(x)\in f(A\cap B)\Rightarrow x\in A\cap B$$ but in my counterexample we have $f(-1)=1\in f(A)$ and $-1\not\in A=(0,\infty)$.
  • The second by writing $$f(x)\in f(A)\land f(x)\in f(B)\Rightarrow x\in A\land x\in B$$ and the counterexample is $1=f(1)\in f(A)\land f(1)\in f(B)$ but $1\not\in B=(-\infty,0)$
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  • $\begingroup$ what if $A\cap B\neq \emptyset$ ... I am sure he mean non empty intersection..... :O $\endgroup$ – user87543 Apr 11 '14 at 14:15
  • $\begingroup$ In your proof there's a mistake: if $y=f(x)\in f(A\cap B)$ then this doesn't mean that $x\in A\cap B$ but that $$x\in f^{-1}(f(A\cap B))$$ and in general we have $$A\cap B\subset f^{-1}(f(A\cap B))$$ @LePressentiment $\endgroup$ – user63181 Apr 11 '14 at 14:30
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    $\begingroup$ @LePressentiment Although you didn't ask for a counterexample, this (or a simpler) counterexample is an excellent tool for answering the question you did ask. Once you have a counterexample, you can insert that specific $f$ and $A$ and $B$ into your attempted proof and see which of the statements are true and which are false. Where you see an $\implies$ pointing from a true to a false statement, that's where your error is. $\endgroup$ – Andreas Blass Apr 11 '14 at 22:04
  • $\begingroup$ @SamiBenRomdhane Thank you. Would you please relocate your comment into your answer? Moreover, how would I continue the string of equivalences from $x\in f^{-1}(f(A\cap B)) $? Will you please to respond in your answer owing to the limitations of comments? $\endgroup$ – Greek - Area 51 Proposal Apr 12 '14 at 6:22
  • $\begingroup$ @AndreasBlass Thank you for the advice. $\endgroup$ – Greek - Area 51 Proposal Apr 12 '14 at 6:23
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The proof above actually shows that $f(A \cap B) \subseteq f(A) \cap f(B)$. The breakdown is that: $$ x \in A \cap B \implies f(x) \in f(A \cap B) $$ but not the other way around. For example, $A = \{0,1\}$, $B = \{-1, 0\}$, $f(x) = x^2$.
In this case, $f(-1) = 1 \in f(A)$, but $-1 \notin A$.

On the other hand, if $f(x)$ is one-to-one, then $$ f(b) = f(x) \implies b = x $$

Also, by definition of $f(A)$, we have: $$ y \in f(A) \implies \exists \, a \in A \mid f(a) = y $$

Substitute $f(x) = y$ and combine these, to get: $$ f(x) \in f(A) \Leftrightarrow x \in A. $$

Thus, if f is one-to-one, then the implication in the first line of your proof is bidirectional, and the proof works.

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  • $\begingroup$ Thank you. Would you please elucidate "substitute $f(x) = y$ and combine these"? What should be substituted and combined? $\endgroup$ – Greek - Area 51 Proposal Apr 12 '14 at 6:19
  • $\begingroup$ Use "f(x)" in place of "y" in the previous equation (also "b" in place of "a"). Then the two equations become $f(x) \in f(A) \implies \exists b \in A | f(b) = f(x) \implies b = x \implies x \in A$. $\endgroup$ – user3294068 Apr 14 '14 at 21:17
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Your problem is that $s \in S$ is not equivalent to $g(s) \in g[S]$. You have made this mistake several times in your derivation.

As a simple counter-example, let $g$ be the function $g(x) = x^2$ on the reals, let $S$ be the set of positive numbers, and let $s = -1$.

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