why is showing that $f(H) \subseteq H$ the same as showing that $(y \in H \implies f(y) \in H)$? 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?
 A: 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 $(\forall y\in H):f(y)\in H$.
A: 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 $y \in H$.
A: 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: $$\begin{align}
f(H)\subseteq H
& \iff \forall x. x\in f(H) \Rightarrow x\in H \\
& \iff \forall x. x\in \{f(y)\mid y\in H\} \Rightarrow x \in H \\
& \iff \forall x. (\exists y.y\in H\land x = f(y) ) \Rightarrow x \in H \\
& \iff \forall x. \forall y. (y\in H\land x = f(y)) \Rightarrow x \in H \\
& \iff \forall x. \forall y. y\in H \Rightarrow f(y) \in H \\
& \iff \forall y. y\in H \Rightarrow f(y) \in H
\end{align}$$
A: 
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:


*

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

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

*$A \subseteq f^{-1}(B)$

*$f(A) \subseteq B$

