How to prove, when $f(A) \subseteq f(B)$ doesn't "always" mean that $$A \subseteq B$$
when $ f\colon X \to Y $ is total function (not partial)
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$\begingroup$ Note that in each of the answers below, we are dealing with functions that are not one-to-one. It's a good exercise to show that a function $f:X\to Y$ is one-to-one if and only if $f(A)\subseteq f(B)$ always means $A\subseteq B$ for $A,B\subseteq X.$ $\endgroup$– Cameron BuieNov 5, 2013 at 14:38
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5 Answers
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take $f$ to be a real valued constant function defined on the set of all real numbers, ie $f(x) = 1$ for all real $x$. then the relation $f(A) \subset f(B)$ is always satisfied
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To prove that an implication isn't true, i.e., that it doesn't always hold, we need only exhibit a counterexample (or example showing when the implication is false).
Consider the function $f(x) = x^2$, $A = \mathbb R,\; B= \mathbb R_{\geq 0}$.
$f(A) = f(B),$ but $A \not\subseteq B$.
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$\begingroup$ i thought about this too, but the problem is in the defination of the total function which says that $\forall x \in X$, $\exists$ exactly one $y \in Y$, in your example $ f(2) = f(-2) = 4 $ \|f^(-1)(4)\| > 1$ $\endgroup$ Nov 5, 2013 at 14:31
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$\begingroup$ I think you're misunderstanding total function vs. partial function. A total function is a function: it maps every $x$ in the domain to some $y$ in the codomain. This does not require that the function be one-to-one, which is what you are implying. $\endgroup$– amWhyNov 5, 2013 at 14:36
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Consider $f: \mathbb R \to \mathbb R^+, x\mapsto x^2$ And then see $$f([0, 2]) = f([-2, 0])$$
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Take $X=\{1,2\}$, $Y=\{y\}$ and $f(1)=f(2)=y$ (no ambiguity). Then $f(\{1\})=f(\{2\})=Y$.
answered Nov 5, 2013 at 14:23
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Just consider the map $f:\mathbb{R}\rightarrow \left\{0\right\}$, and any two disjoint stes in $\mathbb{R}$. For example, consider $I=\left[0,1\right]$ and $J=\left[-1,0\right)$. Clearly $I\cap J=\emptyset$ and $f(I)=\left\{0\right\}\subseteq f(J)=\left\{0\right\}$. Hence, $f$, $I$, and $J$ are a counterexample for the given proof.
