Vtyjkyuk

Question. Pick out the true statements.

1. Let $f:BbbZ to BbbZ^2$ be a bijection. There exists a continuous function from $BbbR$ to $BbbR^2$ which extends $f$.




2. Let $D$ denote the closed unit disc in $BbbR^2$. There exists a continuous mapping $f:Dsetminus (0,0)to x$ which is onto.




3. Let $D$ denote the closed unit disc in $BbbR^2$. Then there exists a continuous mapping $f:D setminus (0,0)to >1$ 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 $BbbR$ on $BbbR^2$ 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 $>1$ is disconnected.)

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

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

$$
beginarraylcl
a &=& lfloor x rfloor \
g(x)&=&
f(a)x+(f(a+1)-f(a)) lbrace x rbrace
endarray
$$

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