Let $S$ denote the set of points on the unit circle centred at $(0,0)$. Does there exist an injective function $f : S \rightarrow S \setminus \{(1,0)\}$?
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The two sets have the same cardinality, so there is even a bijection between them. – Kavi Rama Murthy May 27 '20 at 06:33
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@KaviRamaMurthy Can you construct such a function? – Ananda Bhaduri May 27 '20 at 06:40
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Essentially a duplicate of [How to define a bijection between $(0,1)$ and $(0,1]$?](https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1). – Martin R May 27 '20 at 06:47
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Sure. Just take a injective sequence of points $(x_n)$ on the unit circle, setting $x_1=(1,0)$, and define $f:S^1\to S^1\setminus\{(1,0)\}$ by $f(x)=\begin{cases} x_{j+1}\,,x=x_j\\x\,,x\notin(x_n)\end{cases}$.
For instance, you could let $x_n=e^{2\pi i/n}$.
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Let $v_1=(1,0), v_2=(\cos 1, \sin 1), v_3=(\cos \frac1 2, \sin \frac 1 2 ),v_4=(\cos \frac 1 3, \sin \frac 1 3 ),....$. Define $f(v)=v$ if $f \notin \{v_1,v_2,...\}$, $f(v_1)=v_2,f(v_2)=v_3,...$. This gives a bijection from $S$ onto $S\setminus \{(1,0)\}$.
Kavi Rama Murthy
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Idiotic downvoting of most of my answers these days. Can some one tell me why the other answer is better than mine? – Kavi Rama Murthy May 27 '20 at 07:23