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Let $f : (X, d_1) \rightarrow (Y, d_2)$ be a map between metric spaces and let $T_1$ (and $T_2$ respectively) be the topology on $X$ (and on $Y$ respectively) that is induced by $d_1$ ($d_2$ respectively).

Show using the $\epsilon$-$\delta$ definition of continuity that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous iff $f^{-1}(U)\in T_1$ for all $U\in T_2$.

$\Rightarrow$" :

Suppose that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous.

Given $\epsilon > 0$ there exists $\delta > 0$ such that $$d_1(x, y) < \delta \implies d_2(f(x), f(y)) < \epsilon$$ right?

To show that $f^{-1}(U)\in T_1$ for all $U\in T_2$ we have to show that if $a\in f^{-1}(U)$ then $a\in T_1$ ?

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    $\begingroup$ The condition you first give is uniform continuity, which is stronger than continuity. You instead have that given both $x\in X$ and $\epsilon>0$ there is $\delta>0$ such that $d_1(x,y)<\delta$ implies $d_2(f(x),f(y))<\epsilon$. You will want to show that if $U\in T_2$ and $a\in f^{-1}(U)$ then $f^{-1}(U)$ contains a ball around $a$. For this use that $U$ contains a ball around $f(a)$ and use the $\epsilon-\delta$ definition of continuity. $\endgroup$
    – A Epelde
    Commented Apr 20, 2021 at 16:03
  • $\begingroup$ So $f^{-1}(U)\in T_1$ is equivalent to teh fact that $f^{-1}(U)$ contains a ball around $a$ ? @AlejandroEpelde $\endgroup$
    – Mary Star
    Commented Apr 21, 2021 at 7:52
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    $\begingroup$ Not quite. $f^{-1}(U)\in T_1$ is equivalent to for all $b\in f^{-1}(U)$, $f^{-1}(U)$ containing an open ball around $b$. This is just the definition of the topology coming from a metric. $\endgroup$
    – A Epelde
    Commented Apr 21, 2021 at 7:55
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    $\begingroup$ Members of $T_1$, $T_2$ are called open sets. A set is open iff it is a neighborhood of each of its points. $\endgroup$
    – absolute0
    Commented Apr 21, 2021 at 8:51

1 Answer 1

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Let $f$ is continuous. $U$ be any open set in $Y$. We will show that $f^{-1}(U)$ is open in $X$. Let $a \in f^{-1}(U)$, $f(a) \in U$. Since $U$ is open there exists $\epsilon>0$ such that $B_Y(f(a),\epsilon)\subseteq U$. Now by continuity at $a$, there exists $\delta>0$ such that $f(B_X(a,\delta))\subseteq B_Y(f(a),\epsilon)\subseteq U$. This implies $a \in B_X(a,\delta)\subseteq f^{-1}(U)$. Thus $f^{-1}(U)$ is open in $X$.

Now let $a \in X$. We will show that $f$ is continuous at $a$. For any $\epsilon>0$, $f(a)\in B_Y(f(a),\epsilon)$. $f^{-1}(B_Y(f(a),\epsilon))$ is open in $X$ containing $a$. So there exists $\delta>0$ such that $B_X(a, \delta)\subseteq f^{-1}(B_Y(f(a),\epsilon))$. This implies $f(B_X(a, \delta))\subseteq B_Y(f(a),\epsilon)$. Thus $f$ is continuous at $a$.

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  • $\begingroup$ As for the first part: Why do we need to take that $U$ is open? And why do we have to show that $f^{-1}(U)$ in open in $X$ to get the desired result? Since we defined $U$ to be open we have that every point in $U$ is the center of an open ball contained in $U$, right? By continuity at $a$ we have that $$\forall\epsilon>0 \ \exists \delta>0: d_1 (x,a)<\delta \Rightarrow d_2(f(x),f(a))<\epsilon$$ or not? Is this equivalent to $$\forall \epsilon>0 \ \exists \delta>0 : f(B_X(a,\delta))\subseteq B_Y(f(a),\epsilon)\subseteq U$$ ? $\endgroup$
    – Mary Star
    Commented Apr 21, 2021 at 7:48
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    $\begingroup$ In the first part it's assumed that $f$ is continuous. Then we show that preimage of open set is open, as desired. ( We do the reverse in the second part.) The two expressions you wrote are equivalent. $\endgroup$
    – absolute0
    Commented Apr 21, 2021 at 7:53
  • $\begingroup$ Ok! As for the second part: Why do we take at the beginning $f(a)\in B_Y(f(a),\epsilon)$ ? Do we get that by the assumption of this part, i.e. that the preimage of open set is open? $\endgroup$
    – Mary Star
    Commented Apr 21, 2021 at 13:20
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    $\begingroup$ Here it's shown that $f$ is continuous at $a$. For that it's necessary to get: for any $\epsilon >0~~~\exists~\delta>0$ such that $f(B_X(a, \delta))\subseteq B_Y(f(a),\epsilon)$. That's why the set $B_Y(f(a),\epsilon)$ is considered first. $\endgroup$
    – absolute0
    Commented Apr 21, 2021 at 13:28
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    $\begingroup$ You're welcome. $\endgroup$
    – absolute0
    Commented Apr 22, 2021 at 17:32

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