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A mapping $f : X \rightarrow Y$ is said to be $characteristic$ if for every compact $C \subseteq Y$ the preimage $f^{-1}(C) \subseteq X$ is also compact.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and characteristic. Prove that the $f$ is closed mapping, if the topology on $\mathbb{R}$ is standard topology.

I tried with $A \subseteq \mathbb{R}$ and $f(A) = B$ to prove that $Cl(B) = B$. I couldn't finish it.

Nemanja Beric
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I will replace “characteristic” with “proper”, since it is the standard terminology.

Let $A \subset \mathbb{R}$ be a closed subset, and $B=f(A)$. Let $b_n$ be a sequence of elements of $B$ converging to $b$. Then $K=\{b_n,\,n\} \cup \{b\}$ is compact, so its pre-image $L’$ is compact.

We can write $b_n=f(a_n)$, $a_n \in A$. By definition $a_n \in L$ thus there is a subsequence that converges to $a \in L \cap A$. The corresponding subsequence of $b_n$ goes to $f(a)$, so $b=f(a) \in B$.

Aphelli
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  • How do I know that the $b$, as the limit point of the sequence, even exists? – Nemanja Beric Jan 30 '19 at 20:20
  • Because that is the definition of a closed subset: you take a converging sequence and you show that the limit is in the subset as well. – Aphelli Jan 30 '19 at 20:24
  • if $B = f(A)$, how do I know that the $B$ is closed? That is what I am trying to prove... What am I missing? – Nemanja Beric Jan 30 '19 at 20:25
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    I take $A$ closed and $B=f(A)$. I want to prove that $B$ is closed. So I take a converging sequence of elements of $B$ and then ... and then I conclude that the limit is in $B$. Which proves $B$ to be closed. – Aphelli Jan 30 '19 at 20:27