Does a bijection $f:\Bbb{Z} \to \Bbb{Z}^2$ have an continuous extension from $\Bbb{R}$ to $\Bbb{R}^2$

general-topology

Question. Pick out the true statements.

  1. Let $f:\Bbb{Z} \to \Bbb{Z}^2$ be a bijection. There exists a continuous function from $\Bbb{R}$ to $\Bbb{R}^2$ which extends $f$.

  2. Let $D$ denote the closed unit disc in $\Bbb{R}^2$. There exists a continuous mapping $f:D\setminus \{(0,0)\}\to \{x \in \Bbb{R} \mid |x| \le1\}$ which is onto.

  3. Let $D$ denote the closed unit disc in $\Bbb{R}^2$. Then there exists a continuous mapping $f:D \setminus \{(0,0)\}\to \{x \in \Bbb{R} \mid |x|>1\}$ which is onto.

My Attempt.

  1. For this option try to think geometrically but cannot properly figure out such an extension. (I thought this extension as a long curve on $xy$-plane joining each $f(n)$….I actually try to PASTE $\Bbb{R}$ on $\Bbb{R}^2$ like that curve but pasting each $n$ with its image $f(n)$…but it is not so clear enough too.)

  2. True (The map $(x,y)\mapsto x$ is such a map.)

  3. False ($D \setminus \{(0,0)\}$ is connected whereas $\{x \in \Bbb{R} \mid |x|>1\}$ is disconnected.)

Can anyone please help me in the option 1. Thank you.

Best Answer

To make Lord Shark's comment more explicit define the extension $g$ as

$$ \begin{array}{lcl} a &=& \lfloor x \rfloor \\ g(x)&=& f(a)x+(f(a+1)-f(a)) \lbrace x \rbrace \end{array} $$

where $\lfloor x \rfloor$ is the floor of $x$ and $\lbrace x \rbrace = x - \lfloor x \rfloor$ is the fractional part.

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