4
$\begingroup$

Show that if the sets $\{x:f(x)<a\}$ and the set $\{x:f(x)>a\}$ are open for every $a\in \Bbb Q$ where $(M,d)$ is a metric space and $f:M\to\Bbb R$ ,then $f$ is continuous.

Take a basic open set $(a,b)\in \Bbb R$. To show that $f$ is continuous we need to show that $f^{-1}(a,b)$ is open.

Let $c_n(\in \Bbb Q)\to a$ and $d_n(\in \Bbb Q)\to b$

$f^{-1}(a,b)= \{x:f(x)>a\}\cap \{x:f(x)<b\}$

But how can I write $\{x:f(x)>a\},\{x:f(x)<b\}$ in terms of $c_n,d_n$??

Please help.

$\endgroup$
1
5
+50
$\begingroup$

Let $a\in \mathbb R.$ Then there exists a sequence of rationals $c_1>c_2 > \cdots$ such that $c_n\to a.$ It follows that $(a,\infty) = \cup_{n=1}^\infty (c_n,\infty).$ Thus, since inverse images respect unions, we have

$$\tag 1 \{x: f(x)>a\} = f^{-1}((a,\infty))= f^{-1}(\cup_{n=1}^\infty (c_n,\infty))= \cup_{n=1}^\infty f^{-1}((c_n,\infty)).$$

We are given that each $f^{-1}((c_n,\infty))$ is open. Since any union of open sets is open, the right side of $(1)$ is open.

We have shown $\{x: f(x)>a\}$ is open for every real $a.$ A similar argument shows $\{x: f(x)<b\}$ is open for every real $b.$

Please ask if you have any questions.

$\endgroup$
2
  • $\begingroup$ I got your argument. But there is something wrong in your last line. Because we have to prove $f$ is continuous whether it is given that both sets $\{x:f(x)<a\}$ and $\{x:f(x)>a\}$ are open. $\endgroup$ – Empty Jun 27 '17 at 16:04
  • $\begingroup$ The question was "But how can I write $\{x:f(x)>a\},\{x:f(x)<b\}$ in terms of $c_n,d_n$??" I answered that and more. Rather than saying my last line is wrong, which is false, you could have asked a question. $\endgroup$ – zhw. Jun 27 '17 at 16:16
4
$\begingroup$

Hint: arbitrary unions of open sets are open.

Can you write $\{x:f(x)<a\}$ as a countable union of open sets, where $a\in\mathbb{R}$ $($not just $\mathbb{Q})$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.