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While I was reading a proof about the diagonalizability of symmetric matrices I got a bit lost the author was supposed to show that $f(H) \subseteq H$ but he ended up showing that

$y \in H \implies f(y) \in H$ for all $y$

claiming it's the same thing

I would say because $y$ is arbitrary here $f(y)$ is practically the same as $f(H)$ since $y \in H$ looking at it this way it makes some sense to me but I'm still confused why those two statements are equivalent.

Please could anybody here clearly explain why?

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    $\begingroup$ I do believe that $f(A)$ for some set $A \subseteq D$ where $D$ is the domain of $f$ is defined as the set of all values of $f(a) \space \forall a \in A$. That would make the two statements equivalent. $\endgroup$ – Tim The Enchanter Jun 14 '17 at 10:00
  • $\begingroup$ @TimTheEnchanter thank you ! now I get it. $f(H) = \{f(y) | \forall y \in H \}$ so showing that $y \in H \implies f(y) \in H$ means showing that every element of $f(H)$ is an element of $H$ which is equivalent to $f(H) \subseteq H$ $\endgroup$ – the_firehawk Jun 14 '17 at 10:09
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Here's a systematic, "formal" derivation. The definition of $X \subseteq Y$ is $\forall x.x\in X \Rightarrow x\in Y$. The definition of $f(H)$ is $\{f(y)\mid y\in H\}$ or more explicitly $\{z\mid\exists y.y\in H\land z = f(y)\}$. We can now just calculate: $$\begin{align} f(H)\subseteq H & \iff \forall x. x\in f(H) \Rightarrow x\in H \\ & \iff \forall x. x\in \{f(y)\mid y\in H\} \Rightarrow x \in H \\ & \iff \forall x. (\exists y.y\in H\land x = f(y) ) \Rightarrow x \in H \\ & \iff \forall x. \forall y. (y\in H\land x = f(y)) \Rightarrow x \in H \\ & \iff \forall x. \forall y. y\in H \Rightarrow f(y) \in H \\ & \iff \forall y. y\in H \Rightarrow f(y) \in H \end{align}$$

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I don't know why you wrote that “$f(y)$ is practically the same thing as $f(H)$ since $y\in H$”. It happens that $f(y)$ is the image of $y$, whereas $f(H)$ is the set of all images of all elements of $H$. So, yes, asserting that $f(H)\subseteq H$ actually is the same thing as asserting that $(\forall y\in H):f(y)\in H$.

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Saying $f(H) \subseteq H$ is saying that $f$ takes each element of $H$ into $H$. Thus for all $y \in H$, $f(y) \in H$ and the statement $y \in H \Rightarrow f(y) \in H$ holds.

Conversely, if the implication $y \in H \Rightarrow f(y) \in H$ holds, $f(H) \subseteq H$ becuase by the implication $f(y) \in H$ for all $y \in H$.

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Proposition 0. Given a function $f : X \rightarrow Y$ and subsets $A \subseteq X, B \subseteq Y$, the following are equivalent:

  • $\forall x \in X(x \in A \rightarrow f(x) \in B)$
  • $f(A) \subseteq B$

Proof.

The following are equivalent:

  • $\forall x \in X(x \in A \rightarrow f(x) \in B)$
  • $\forall x \in X(x \in A \rightarrow x \in f^{-1}(B))$
  • $A \subseteq f^{-1}(B)$
  • $f(A) \subseteq B$
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