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How to show that $(1)\Longleftrightarrow (2)$ in metric spaces ?

  1. pre-image of open sets are open
  2. $\delta$-$\epsilon$ definition of continuity
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up vote 5 down vote accepted

(1)$\Rightarrow$(2) Let $f\colon (X,d_X)\to (Y,d_Y)$ be a function of metric spaces such that for every open subset $U$ of $Y$, $f^{-1}(U)$ is open on $X$.

Let $x\in X$, and let $\epsilon\gt 0$. We want to show that there exists $\delta\gt 0$ such that $d_X(a,x)\lt \delta$ implies $d_Y(f(a),f(x))\lt \epsilon$. Think about the open set $$B(f(x),\epsilon) = \{y\in y\mid d_Y(y,f(x))\lt\epsilon\}$$ and about the definition of "open set" in $(X,d_X)$.

(2)$\Rightarrow$(1). Let $f\colon (X,d_X)\to (Y,d_Y)$ be a function of metric spaces such that for every $x\in X$ and every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $d_X(a,x)\lt \delta$, then $d_Y(f(a),f(x))\lt\epsilon$. We need to show that if $U$ is an open subset of $Y$, then $f^{-1}(U)$ is an open subset of $X$.

Let $U$ be an open subset of $Y$, and let $x\in f^{-1}(U)$. Since $f(x)\in U$ and $U$ is open, there is an open ball $B(f(x),\epsilon)$, $\epsilon\gt 0$, such that $B(f(x),\epsilon)\subseteq U$. Then...

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Hmm, what if the pre-image of B(f(x),eps) is unbounded? eg. the positive real line – Laylady Feb 3 '12 at 2:11
@Laylady: What is it that you need to prove? You need to prove that $f^{-1}(U)$ is open. What does that mean? It means that around every point $x\in f^{-1}(U)$, you can find an open ball centered at $x$ that is completely contained in $f^{-1}(U)$. What does it mean for a subset of $X$ to be completely contained in $f^{-1}(U)$? That when you apply $f$, you "end up" in $U$. Whether the pre-image of $B(f(x),\epsilon)$ is bounded or unbounded is completely irrelevant and immaterial to any of this. – Arturo Magidin Feb 3 '12 at 4:37

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