the problem is Let A = {1/n : n ∈ N} and π : I → I/A be the projection map. Let G = [0, 1/2] ∪ {1}. Show that G is a saturated nbd of 0, but π(G) is not a nbd of π(0).
The relationship here would be the elements of G that are in A? Any suggestion or clue
Let $A=\{1/n : n \in \mathbb{Z}_+ \},$ $I=[0,1].$ The space $I/A$ is the quotient space that collapses the set $A$ in a single point. Let $a \in I/A$ be the point in which $A$ is collapsed. Then $$ I/A = \{a \} \cup \{ \{x \} : x \in I, x \neq 1/n \text{ for }n \in \mathbb{Z}_+\}.$$
Let $p:I \to I/A$ be the quotient map. Let $G=[0,1/2]\cup \{1\}.$ To show that $G$ is saturated, we must show it is the complete inverse image of a subset of $I/A.$ For this, try $$B=\{a\}\cup\{\{x\}: x \in I, x \neq 1/n \text{ for }n \in \mathbb{Z}_+, 0 \leq x < 1/2\}.$$ Then $B \subset I/A$ and $G=p^{-1}(B).$
To show that $\pi(G)$ is not a neighborhood of $\pi(0),$ note that $p(G)=p(p^{-1}(B))=B$ ($p$ is surjective), but $B$ is not open in $I/A$ since $\pi^{-1}(G)$ is not open in $I.$ (Definition of quotient topology.)
Note that a quotient map maps saturated open sets of $I$ to open sets of $Y,$ but this is not the case since $G$ is not open in $I.$