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If $p(t)=(1-t)a+tb$ with $a∈A$, $b∈B $ and $t∈R^1$, how can I prove that $A_0=p^{-1}(A)$ and $B_0=p^{-1} (B) $ are separated subsets of $R^1$?

Remark: in a metric space two set A and B are said to be separated if $A∩{\overline B} =∅$ and ${\overline A}∩B=∅$, this means that no point of A belongs to the closure of B and no point of B belongs to the closure of A.

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    What are they separated *by*? If $A$ and $B$ are separated by some set $E$ (e.g., a hyperplane), then $p^{-1}(E)$ will provide a nice separator. – 2017-01-10
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    What is separated? Strictly separated by a hyperplane? – 2017-01-10
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    Separated in the context of metric spaces – 2017-01-10
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    ...Or, plainly, separated in the sense of disjoint sets, i.e. with no common point : $A \cap B=\emptyset$ ? – 2017-01-10
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    Hint: First show that for a continuous function the closure of the preimage is contained in the preimage of the closure. – 2017-01-10
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    t is in R, or (0,1)? – 2017-01-12
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    t belongs to R.... – 2017-01-12
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    @DIEGORAMOS: Linear functions such as $p(t)$ are continuous. – 2017-01-12
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    @MoisheCohen :Yes, this exercise was proposed in a exam, where nothing about continuity is known – 2017-01-13
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    @DIEGORAMOS: It is hard to imagine that you are taking this class before taking calculus classes/real analysis, where continuity of linear functions is at least stated (calculus)/proven (real analysis). My guess is that whoever wrote this exam assumed that you already know that linear functions are continuous. – 2017-01-13

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