Question: Every bijection $f:\Bbb{R}\rightarrow[0,\infty)$ has infinitely many points of discontinuity.
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You may refer this - https://math.stackexchange.com/q/8149/802621 – Ayush Jain Jun 23 '20 at 07:48
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The following is a topological insight. I will show there is at least one discontinuity.
If possible let $f:\Bbb R\to [0,\infty)$ be a bijective continuous function. Then, one can show $\lim_{x\to \pm \infty}f(x)=\infty$. In other words, we can extend the map $f$ to a map $\widetilde f:\Bbb S^1\to [0,\infty]$. Note that $\Bbb S^1$ is the one point compactification of $\Bbb R$ and $[0,\infty]\simeq [0,1]$ is one point compactification of $[0,\infty)$. Also, $\widetilde f$ is bijective, hence a homeomorphism. But we know, $\Bbb S^1$ is not homeomorphic to $[0,1]$.
Sumanta
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From a compact space to a Hausdorff space a bijective map is always is homeomorphism. – Sumanta Jun 23 '20 at 08:20
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