Is there a continuous injective map from $\mathbb{R}$ that has compact image?

Question

Suppose I have a function $f: \mathbb{R} \to X$, where $X$ is some non-compact metric space.

Is it possible that $f$ is injective yet has compact image? If the answer is yes, what characterizes those cases? What additional properties would make the answer negative?

My progress so far: If the image is compact, $f(n)$ has a convergent subsequence with limit in the image, i.e. $f(a_n) \to f(a)$ where $a_n \to \infty$. But I find it tricky to turn the limit into equality.

score 9 · Accepted Answer · answered Feb 15 '14 at 21:38

9

Take $X = \mathbb{R}^2$ and $f$ an immersion of $\mathbb{R}$ which makes a figure eight:

enter image description here

answered Feb 15 '14 at 21:38

Neal

34,012

Huh, that's pretty much the same as what I thought, only much simpler... – tomasz Feb 15 '14 at 21:38
Ah, okay, with the image, I see it more clearly – Thomas Andrews Feb 15 '14 at 21:40
1

Then again, mine can be done without diminishing derivative, so I guess it still says something. :) – tomasz Feb 15 '14 at 21:47

tomasz · Answer 2 · 2014-02-15T21:42:52.040

4

Yes, it is possible. Consider a function $f\colon {\bf R}\to {\bf R}^2$ which takes the line to a doubled topologist's sine curve, looping around through the central interval.

This map can be made to be differentiable, and an arc length parametrisation, so requiring this property won't make the answer negative (not even in addition to requiring $X$ to be euclidean).

edited Feb 15 '14 at 21:42

answered Feb 15 '14 at 21:37

tomasz

37,896

Is there a continuous injective map from $\mathbb{R}$ that has compact image?

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