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Does there exist an embedding $f: \mathbb{R} \rightarrow S^{2}$ with a closed image? I believe not, but I'm stuck with how to prove that.

It would be nice to hear several different proofs if my guess is true.

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    $\begingroup$ The term "embedding" (in the question title, but not in the question body) usually means "$f$ is a homeomorphism onto its image". Just checking: Do you want that condition, or not? $\endgroup$ – Andrew D. Hwang Apr 9 '15 at 10:42
  • $\begingroup$ Try to use the claim known as the Invariance of Domain. Consequently this fact $f$ must be surjective too. (if $f$ is a topological embedding) $\endgroup$ – Leonhardt von M Apr 9 '15 at 10:46
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    $\begingroup$ @user86418 I want an embedding, yes, thanks. $\endgroup$ – lisyarus Apr 9 '15 at 10:48
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Let $f: \mathbb R \to S^1$ topological embedding and let $R=f(\mathbb R)$. If $R$ is closed in $S^1$ $R$ is compact and $f$ is a homeomorphism between a compact and a non compact topological space. This is a contradiction.

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  • $\begingroup$ Couldn't image it to be that simple. Thank you! $\endgroup$ – lisyarus Apr 9 '15 at 11:46

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