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.
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 $b=f(a) \in B$.