# Show that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous iff $f^{-1}(U)\in T_1$ for all $U\in T_2$

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 $$a\in T_1$$ ?

• 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. Commented Apr 20, 2021 at 16:03
• So $f^{-1}(U)\in T_1$ is equivalent to teh fact that $f^{-1}(U)$ contains a ball around $a$ ? @AlejandroEpelde Commented Apr 21, 2021 at 7:52
• 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. Commented Apr 21, 2021 at 7:55
• Members of $T_1$, $T_2$ are called open sets. A set is open iff it is a neighborhood of each of its points. Commented Apr 21, 2021 at 8:51

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 $$a$$.
• 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 $$\forall \epsilon>0 \ \exists \delta>0 : f(B_X(a,\delta))\subseteq B_Y(f(a),\epsilon)\subseteq U$$ ? Commented Apr 21, 2021 at 7:48
• 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. Commented Apr 21, 2021 at 7:53
• 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? Commented Apr 21, 2021 at 13:20
• 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. Commented Apr 21, 2021 at 13:28