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

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.)

• 1. Linearly interpolate. Sep 1 '18 at 7:05
• @LordSharktheUnknown....well... what is the role of the bijection $f$ in linear interpolation....? Sep 1 '18 at 7:09
• Absolutely none! Any map $\Bbb Z\to\Bbb R^2$ can be so interpolated. Sep 1 '18 at 7:09

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.
• But $g$ should have $\Bbb{R}$ as domain...!!! Sep 1 '18 at 7:22
• @IndrajitGhosh It has domain $\mathbb R$ now. Sorry for the confusion in my initial answer. Sep 1 '18 at 7:25