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Let $f:\mathbb R \to \mathbb R$ be a function such that it carries compact sets to compact sets and $f^{-1}(\{x\})$ is closed for every $x \in \mathbb R$ , then is $f$ continuous ?

(I know that if $f$ is a function on real line having ivt and preimage of every singleton is closed then $f$ is continuous ; also if a function on real line has ivt and carries compact sets to compact sets then $f$ is continuous ; this question is motivated from theses two facts )

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Suppose $f$ is not continuous from the right at $x_0$ and $f(x_0)=y_0$. Notice that for all $x>x_0$, $f([x_0,x])$ is compact. Then there exists a sequence $\{x_n\}_{n\in\mathbb{N}}$ and $y_1\neq y_0$ such that $x_n>x_0$, $\lim_{n\to+\infty}x_n=x_0$ and $\lim_{n\to+\infty}f(x_n)=y_1$. Furthermore, $y_1\in f([x_0,x])$ for all $x$ because of the compactness. Also $f^{-1}(\{y_1\})$ is closed, and $f(x_0)=y_0\neq y_1$. So $x_0$ lies in an open set $(a,b)$ such that $y_1\notin f((a,b))$. It's a contradiction.

lostlife
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  • Okay , so the limit of $f(x_n)$ exists ( actually limit of a subsequence but we can rename anyway) as $f(x_n)$ is a sequence in the compact set $f[x_0,x]$ –  Jun 18 '16 at 08:45
  • How do you know $y_1 \in f[x_0,x]$ for every $x>x_0$ ? Your $y_1$ should depend on $x$ : initially you just have a sequence $x_n>x_0$ with $x_n \to x_0$ ; you don't know whether $\lim f(x_n)$ exists or not ; then you choose some fixed $a>x_0$ , as $x_n \to x_0$ and $x_n >x_0$ so $x_n \in [x_0,a]$ ultimately , so there is a natural number $k$ s.t. $f(x_n) \in f[x_0,a],\forall n>k $ ; as $f[x_0,a]$ is compact so we have a subsequence $f(x_{r_n})$ in $f[x_0,a]$ such that $y_1=lim f(x_{r_n}) \in f[x_0,a]$ where $r_n>k$ . Now if $x\ge a>x_0$ then $y_1 \in f[x_0,x]$ –  Jun 18 '16 at 11:03
  • so let $a>x>x_0$ ; as $x_{r_n}$ is a subsequence of $x_n$ so $x_{r_n} \to x_0$ and $x_{r_n} > x_0$ so there is natural number $m$ such that $x_{r_n} \in [x_0,x] , \forall n>m$ , i.e. the subsequence $x_{r_n}$ belongs to $[x_0,x]$ ultimately so $f(x_{r_n}) \in f[x_0,x]$ ultimately , thus $y_1=\lim f(x_{r_n})\in f[x_0,x]$ for any $x \in (a,x_0)$ ; This should be the full argument . –  Jun 18 '16 at 11:03
  • The rest of your argument is fantastic , :) +1 –  Jun 18 '16 at 11:12
  • $y_1$ is fixed for all $x\geq x_0$. – lostlife Jun 18 '16 at 14:17
  • That is true but not at all obvious , as indicated in my comments ; because if the condition $f$ carries compact sets to compact sets weren't there , you wouldn't have that $f(x_n)$ converges –  Jun 18 '16 at 14:21
  • Oh, sorry. I thought it is obvious. So I didn't say it clearly. – lostlife Jun 18 '16 at 14:22
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Hint: By the Heine-Borel theorem, every compact subset of $\mathbb R$ is closed and bounded. Therefore, f carries closed sets into closed sets.

Claim: A function f from X into Y is continuous if and only if preimages of closed sets in Y are closed in X.

If you can prove this,this will show the answer is yes.

Mathemagician1234
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  • How come $f$ carries closed sets to closed sets ? I never said $f$ is a closed map .... –  Jun 18 '16 at 08:37
  • If f carries compact subsets to compact subsets of $\mathbb R$ , then by the Heine Borel Theorem, f carries closed and bounded subsets to closed and bounded subsets. Right? – Mathemagician1234 Jun 18 '16 at 15:42
  • Yes right , I have no problem with that , from that it also follows that $f$ carries bounded sets to bounded sets , but how come $f$ becomes a closed map ? –  Jun 18 '16 at 15:56