$f$ is continuous in $\mathbb{R}$ if and only if for every open set $G$, the set $f^{-1}(G)=\{x:f(x)\in G\} $ is open.

Question

Suppose $f(x)$ defined in $\mathbb{R}$, then $f$ is continuous in $\mathbb{R}$ if and only if for every open set $G$, the set $f^{-1}(G)=\{x:f(x)\in G\} $ is open.

I have done necessity.
If $f$ is continuous and G is open. Then for $x_0\in f^{-1}(G)$,$f(x_0)\in G$,and $f$ is continuous hence $\exists U(x_0,\delta)$,such that for $x\in U(x_0,\delta)$,$f(x)\in G$. Thus $ U(x_0,\delta)\subset f^{-1}(G)$ which makes $f^{-1}(G)$ is an open set.
And now I have no ideal of the sufficiency.

To show f is continuous what are you allowed to use? only $\epsilon - \delta$? — gd1035, Jul 17 '18 at 21:18
@ gd1035 If possible, $\epsilon-\delta$ is good for me to understand. — Jaqen Chou, Jul 17 '18 at 21:38

score 2 · Accepted Answer · answered Jul 17 '18 at 21:21

2

Hint: Fix $x \in \mathbb R$. We'll show that $f$ is continuous at $x$. So, fix your favorite $\epsilon > 0$ and consider the open set $(f(x) - \epsilon, f(x) + \epsilon)$. Its preimage under $f$ is open and has the elements $x$. Hence it contains an open interval around $x$. But now the (pointwise) image of that open interval...

answered Jul 17 '18 at 21:21

Stefan Mesken

16,651

the image of the open interval contains $x$ subset $(f(x)-\epsilon,f(x)+\epsilon)$. – Jaqen Chou Jul 17 '18 at 21:36
@JaqenChou You need to show that the $\epsilon$-$\delta$ criterion for continuity at $x$ is satisfied to finish my proof. So... what's $\delta$ here? – Stefan Mesken Jul 17 '18 at 21:40
can we just say that there exists a open interval $B(x,\delta)$ subset the preimage since the preimage is open and has the element x. – Jaqen Chou Jul 17 '18 at 21:50
@JaqenChou Yep. And that $\delta$ will work. (Seems to me like you already know this -- I'm simply stressing that point to make sure you've thought through it.) – Stefan Mesken Jul 17 '18 at 21:52
good teacher of you~ Memeda – Jaqen Chou Jul 17 '18 at 21:54

score 0 · Answer 2 · answered Jul 17 '18 at 21:55

If $x_0 \in X$ and $\varepsilon>0$ the set $G = B(f(x_0), \varepsilon)$ is open. So $f^{-1}[G]$ is open and $x_0 \in f^{-1}[G]$ as $f(x_0) \in G = B(f(x_0), \varepsilon)$ clearly.

So there is some $\delta>0$ such that $B(x_0, \delta) \subseteq f^{-1}[G]$ and it's easy to see that this $\delta>0$ is as required for our starting $\varepsilon$. So $f$ is continuous at $x_0$.

$f$ is continuous in $\mathbb{R}$ if and only if for every open set $G$, the set $f^{-1}(G)=\{x:f(x)\in G\} $ is open.

2 Answers2