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For each $n\in \mathbb{N}$ define:

$B_n := \{(x,y) \in \mathbb{R^2} | (x-n)^2+y^2 \leq \frac{1}{9}\}\quad$ (i.e closed ball around $(n,0)$ with radius $\frac{1}{3}$).

Let $C :=\{(x,y) \in \mathbb{R^2} | y\leq -1\}$

(a) Prove that there exists a function $f : \mathbb{R^2}\rightarrow \mathbb{R}$ such that $f(x,y)=n$ for each $(x,y) \in B_n$ and $f(x,y)=0$ for each $(x,y) \in C$.

(b) Can $f$ be extended to a continuous function on the one-point compactification $\mathbb{R}\cup \{\infty\} $?

For (a) I have tried using Urysohn by observing that, for each $n \in \mathbb{N}$ there is a continuous function $f_n : \mathbb{R^2} \rightarrow [0,1]$ with $f_n|_{B_n} = 1$ and $f_n|_{C} = 0$ and then combining them somehow into a series that yields the desired function (I played around with this idea and the value $1$, even considering the unions of finite closed sets out of $\{B_n\}_{n\in \mathbb{N}}$ and $C$, but this doesn't seem to be going anywhere (in particular, I cannot get it to be $n$ on $B_n$, and/or to converge at all).

I then took a look at Tietze's extension proof again for inspiration, since I realized my attempts at using Urysohn didn't utilize some of the information (the radius and other values), and the Tieze proof uses similar values, but I'm not sure now that this is too relevant.

For (b), if the function discovered is such that it is defined on a closed set in this new space, then we can use Tietze to claim that there is such an extension. Otherwise I'm also not too sure what to do here.

Any advice?

Anon
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1 Answers1

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$\{B_n\mid n \in \Bbb N\} \cup \{C\}$ forms a family of closed sets that is discrete and so the function $f$ that has the required constant values on each of these sets is also continuous and the union of all these sets is closed. So Tietze applies straight away to $f$ and gives the required function on $\Bbb R^2$. For continuity on the $f$ that is partially defined, see e.g. this question, noting that a discrete family, as we have here, is locally finite in particular.

Henno Brandsma
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  • Can you please explain "the function $f$ that has the required constant values on each of these sets is also continuous"? How is this function defined? Perhaps you're referring to a statement as in: https://math.stackexchange.com/questions/704517/every-function-from-a-discrete-subset-is-continuous – Anon Oct 03 '21 at 05:36
  • and then, setting $f:{B_n} \cup C \rightarrow \mathbb{R}$ with a simple $f(x,y)=n$ if $(x,y) \in B_n$ and $f(x,y)=0$ if $(x,y) \in C$ to obtain, for example, that if $U=(0,2)\subset \mathbb{R}$ then $f^{-1}(U)=B_1$ is open in the induced topology. Is that what you mean? – Anon Oct 03 '21 at 06:00
  • @Anon yes, the first $f$ is just $n$ on all points of $B_n$ and $0$ on $C$. This $f$ on the union is continuous by a standard result on continuity of functions defined more generally on locally finite families of closed sets. Etc. – Henno Brandsma Oct 03 '21 at 06:50
  • @Anon See 2 quoted in this question for that result. – Henno Brandsma Oct 03 '21 at 08:14