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Does there exists a continuous onto function from $[0,1)$ to $(0,1)$?

My approach: If there is a continuous onto function from $[0,1)$ to $(0,1)$ then for g , extension of f on [0,1] image of [0,1] is either $(0,1)$ or $[0,1)$ or $(0,1]$ .All of these three sets are non compact .but image of compact set under cont map is compact .

So, it is not possible. Am I right? If Ans is yes then pls help me how to construct such function and how can I visualise it?

Nope
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1 Answers1

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First of all, your explanation are hard to understand. You argument is not true : you cannot always extend continuously to $[0,1]$ a continuous function defined on $]0,1[.$ Indeed, you can think of the map $x \mapsto \frac{1}{x}$.

To answer your question : Does there exists a continuous onto function from [0,1[ to [0,1]?

If we consider the function $$f : [0,1[ \to \mathbb{R} : x \mapsto \frac{x\sin(\frac{1}{x - 1})}{2} +\frac{1}{2},$$ then $f$ is a surjective continuous function from [0,1[ to ]0,1[ as requested. Here is a graph of the function $f$ for intuition.Graph of function <span class=$f$" /> Be careful, the graph is shifted of $0.5$ on the $y-$axis ! This is a mistake.

Cactus
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