# continuous function from $D\setminus \{0\}$ to subset of $\mathbb{R}$

a) is not true as $\mathbb{Z}$ is not dense in $\mathbb{R}$.

b) is not correct as inverse image of a closed set is going to be open set.

c) is not true as $f(D\setminus\{0\})$=is not a connected set.

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I'd appreciate mentioning the source of this problem. –  Martin Sleziak Jul 21 '12 at 9:28

In the case of (a), you say that the function does not extend because $\mathbb Z$ is not dense. However, the density of the subset is only required for the extension to be unique. Think of it this way: the more points you specify, the harder it is to fit a function to them. If you specify on a dense set, you've so many values, you're left with no choice as to what happens elsewhere.

Note that specifying values on a dense set doesn't mean there is an extension, only that it is unique if it exists: the function $$f(x)=\begin{cases}0 &\mbox{if }x^2 < 2 \\ 1 &\mbox{if }x^2 > 2\end{cases}$$ is continuous on $\mathbb Q$ but has no continuous extension to $\mathbb R$. By contrast, given a function on a discrete set like $\mathbb Z$ (i.e. very much smaller than dense!) you can always extend because you've got space around every point in which you can do whatever you like. In fact, it is sufficient (though not necessary) that your set is closed: see the Tietze extension theorem.

Your reasoning in (b) fails because, although you correctly identify that there is a closed set with an open pre-image, this is not in fact a problem because closed sets can also be open and vice versa. In particular, the whole space ($[0,1]$ or $D\setminus\{(0,0)\}$) is always both open and closed.

Your reasoning in (c) is correct.

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For (a), it seems to me that you could construct such a function easily, even though $\Bbb Z$ is not dense in $\Bbb R$ - linear interpolation would be one way to do this.

(b) is certainly true - just map $(x,y)$ to $x$.

I think you're right about (c). It is certainly untrue, for the reason that you gave.

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for the record: I did not downvote you. (as you can see from my profile, I do not downvote) –  Rudy the Reindeer Jul 21 '12 at 9:12
For (c), what about $(r,\theta)\to(1/(1-r),\theta)$? –  Gerry Myerson Jul 21 '12 at 9:17
@MattN. - I don't really care who downvoted me. I'd be more interested in the REASON the unknown person downvoted me; if there's something wrong with my answer, I'd like to fix it. Sadly, this was another hit-and-run. –  user22805 Jul 21 '12 at 9:18
I withdraw my objection, and will look into getting a new pair of glasses. –  Gerry Myerson Jul 21 '12 at 9:30
I'm sure. I'm just prone to understatement. –  user22805 Jul 21 '12 at 11:03

(a) is true. If $x=n+\alpha$, where $0\le\alpha<1$, send $x$ to $\alpha f(n+1)+(1-\alpha)f(n)$.

(b) is true: map $\langle x,y\rangle\in D\setminus\{\langle 0,0\rangle\}$ to $x$.

(c) is false: $D\setminus\{\langle 0,0\rangle\}$ is connected, and $\{x\in\Bbb R:|x|>1\}$ is not.

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@David: You’re right: I misread it. –  Brian M. Scott Jul 21 '12 at 9:19