# Continuous Function + open set

A mapping $T$ of a metric space $X$ into a metric space $Y$ is continuous iff the inverse image of any open subset $Y$ is open subset of $X$.

Proof: (a)Suppose that $T$ is continuous. Let $S \subset Y$ be open and $S_0$ the inverse image of $S$. If $S_0 = \emptyset$, it is open. Let $S_0 \neq \emptyset$. For any $x_0\in S_0$ let $y_0=Tx_0$. Since $S$ is open, it contains an $\epsilon$-neighborhood $N$ of $y_0$. Since $T$ is continuous, $x_0$ has a $\delta$-neighborhood $N_0$ which is mapped into $N$. Since $N\subset S$, we have $N_0\subset S_0$ ...

Please You will be able explain me in detailed this part "Since $N\subset S$, we have $N_0\subset S_0$", I don't understand "we have $N_0\subset S_0$".

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I've changed the word "null" to the symbol of empty set because it was what it seemed you intended to write. So please, verify if I didn't change the meaning of your question. Good luck! – user1620696 May 18 '13 at 3:58
What is your definition of continuous? – Pedro Tamaroff May 18 '13 at 13:16

If $x \in N_0$, then $T(x) \in N$ (as "$N_0$ is mapped into $N$") so $T(x)$ is in $S$ (as we assume that $N \subset S$). So by definition, $x \in S_0$, as $S_0$ is defined the be the inverse image of $S$ ($S_0 = T^{-1}[S]$), so exactly the set of all $p$ with $T(p) \in S$, and the $x$ we start with is such a point.

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When you say: "$T(x) \in S_0" is "T^{-1}(x) \in S_0$"? – juaninf May 18 '13 at 14:19
@juaninf no, I mean $x \in S_0$ (corrected it) – Henno Brandsma May 19 '13 at 8:18

Inverse images preserve set inclusion. More explicitly:

$x\in N_0 \Rightarrow Tx \in N \Rightarrow Tx \in S \Rightarrow x \in S_0.$

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but Why $Tx \in S \Rightarrow x \in S_0$? – juaninf May 18 '13 at 11:22
Which is the guarantee that such for all $\epsilon$ the $\delta$-neighborhood belong $S_0$? – juaninf May 18 '13 at 11:31
Why the inverse image preserver set inclusion? – juaninf May 18 '13 at 11:40