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I am somehow unable to understand the following relation.

if $f:X\rightarrow Y$ and $S,T\subseteq X$ then $$f(S\cap T)\subseteq f(S)\cap f(T)$$

The main problem is I always end up in showing the equality rather than the subset-ness.

Example $f:X=\{x_1,x_2,x_3\}\rightarrow Y=\{y_1,y_2,y_3\}$ where $f$ is defined by $f(x_1)=f(x_2)=y_1$,$f(x_3)=y_3$ and $S=\{x_1\},T=\{x_2,x_3\}$ Then clearly $f(S)\cap f(T)\subseteq f(S\cap T)$ does not hold,thus the equality is not possible.

But My question is where I am making mistake in following proof :

$$y\in f(S) \cap f(T)$$ $$\implies y\in f(S) \land y\in f(T)$$ $$\implies \exists x\in S \land \exists x\in T$$ such that $y=f(x)$ $$\implies x\in S\cap T$$ $$\implies f(x)=y\in f(S\cap T)$$ Hence $$f(S)\cap f(T)\subseteq f(S\cap T)$$

Can you please correct me,?

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    $\begingroup$ The fact that $\exists x \in S \text { and } \exists x \in T$ does not mean that they are the same $x$. $\endgroup$ – Mauro ALLEGRANZA May 16 '19 at 18:23
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    $\begingroup$ Consider : "there is a number that is Even and there is a number that is Odd". $\endgroup$ – Mauro ALLEGRANZA May 16 '19 at 18:24
  • $\begingroup$ Does that mean if such $x$ exists then equality holds ? $\endgroup$ – NewBornMATH May 16 '19 at 18:51
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In line three, you use the same $x$ in both $S$ and $T$. All that you know is that there exists an $x\in S$ and there exists a $z\in T$ so that $f(x)=y=f(z)$. There is no reason that you need the same element in both $S$ and $T$, just a pair of elements with the same image.

In your example, $f(S)=\{y_1\}$ and $f(T)=\{y_1,y_2\}$, so $f(S)\cap f(T)=\{y_1\}$. Then, when you take preimages in $S$ and $T$ of $y_1$, you get $x_1\in S$ and $x_2\in T$. These are (of course) not the same element.

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  • $\begingroup$ (+1) one thing I want to know whether $f(s)$ is the set of exactly all elements that are mapped from $x\in S$ ? I mean consider here f(S)={y_1}, what I want to know , is whether {y1,y2} can be f(S) or not,? (as it contains {y1} !) $\endgroup$ – NewBornMATH May 16 '19 at 18:47
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    $\begingroup$ This is a different question, so it would be better to start a new question if you want more details. However, the short answer is $f(S)$ is exactly the set of images of points in $S$. In other words, $f(S)=\{f(x):x\in S\}$. $\endgroup$ – Michael Burr May 16 '19 at 19:04
  • $\begingroup$ @Michael if it is exactly the set of images then, dpes $f(x)\in f(S)$ necessary imply $x\in S$ ? (Like $x\in S \implies f(x) \in S$) $\endgroup$ – M Desmond May 16 '19 at 19:23
  • $\begingroup$ @MDesmond This is a different question, so it would be better to start a new question if you want more details. The short answer is that the two things you describe are not the same. No, if you know that $f(x)\in f(S)$, then you cannot conclude that $x\in S$ since there might be some other $y\in S$ so that $f(x)=f(y)$. For example, if $S$ is the positive integers and $f$ is the squaring map, then $f(-2)=4\in f(S)$ since $f(2)=4$ and $2\in S$, but $-2\not\in S$. On the other hand, the last line you write is correct (after correcting a typo), $x\in S$ implies that $f(x)\in f(S)$. $\endgroup$ – Michael Burr May 16 '19 at 21:50

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