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The open interval $(-\dfrac{\pi}{2}, \dfrac{\pi}{2})$ is well known to be homeomorphic to $\mathbb{R}$ through the homemorphism $f(x) = \tan(x)$

Is the closed interval also homeomorphic to the real line?

It seems to be the case until you hits the end points, which is undefined by $f(x) = \tan(x)$ so that seems to ruin everything.

Also, is the closed interval homeomorphic to open interval given any intervals?

Shamisen Expert
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    No, it's compact and the reals are not, nor is any open interval. Non-empty intervals all have the same cardinality though. – Noah Olander Jan 17 '16 at 21:42

2 Answers2

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A closed (or half-open) interval is not homeomorphic to the real line.

To wit, a closed interval contains a point that can be removed without making the rest of the space disconnected. The real line doesn't.

(This prevents the two spaces from being homeomorphic because the property of "a subset being disconnected" can be defined in terms only of the topology, which is exactly what a homeomorphism preserves).

hmakholm left over Monica
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  • But can we prove that we cannot construct a homeomorphism? For example, the function $f(x) = \tan(x)$ does not work but it does not exclude the possibility that we can find some other function that it works. Assume I am an idiot. – Shamisen Expert Jan 17 '16 at 21:43
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    @FemaleTank The answer proves there cannot be a homeomorphism. Number of connected components is a homeomorphism invariant. If there is a homeomorphism $f : [a, b] \to \Bbb R$, then $f$ restricted to $[a, b)$ gives a homeomorphism between $[a, b)$ and $\Bbb R - f(b)$. The former is connected, and the latter is disconnected ($f(b)$ is a point). Impossible. – Balarka Sen Jan 17 '16 at 21:45
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No. Continuous functions take compact sets to compact sets. So, there is no continuous function from $[-\pi/2,\pi/2]$ onto $\mathbb R$, since $[-\pi/2,\pi/2]$ is compact, and $\mathbb R$ is not.

Tim Raczkowski
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