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Here is my question.

Let $X=Y=\Bbb R$, with the usual topology. Let $A=[0,1]$ and topologize $A$ with the induced topology from $X$.

Does there exist a continuous function from the topological space $A$ onto $Y$?

Why or why not?

I'd argue No, since we'd be going from a closed to open set. But I also thought that any subset of a topology is an open subset.

Any help would be appreciated.

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  • $\begingroup$ Hint: All of $\mathbb{R}$ is a closed set (also an open set; it's clopen). The tan and arctangent functions have the required type of behavior, just rescale them as needed (domain restrict the tan function to make it invertible). $\endgroup$
    – Justin Benfield
    May 9, 2016 at 12:55
  • $\begingroup$ @JustinBenfield That works for the open interval $(0,1)$, not the closed interval $[0,1]$. $\endgroup$
    – Akiva Weinberger
    May 9, 2016 at 12:56

2 Answers 2

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No: $[0,1]$ is compact, $\mathbb R$ is not compact, and the image of a compact set under a continuous function is always compact.

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The extreme value theorem, from real analysis, states that any continuous function from a compact set to the reals is bounded (and has a maximum). A bounded function cannot be surjective onto $\Bbb R$.

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