In $\mathbb{R}^{\mathbb{R}}$, $E=\{f\in \mathbb{R}^{\mathbb{R}}: f(x)=0$ or $1, f(x)=0 $ only finitely often$\}$

Let g be a function in $\mathbb{R}^{\mathbb{R}}$, and g is identically $0$, then in the product topology of $\mathbb{R}^{\mathbb{R}}$, $g\in Cl(E)$.

How do I find a net $f_n$ converges to g?

I constructed a neighborhood base of g, define a relation on it, and show that it is a directed set, what should I do next?

Thanks for any help!


I’ve a feeling that you may be letting the details get in the way of the basic idea, because once you’ve constructed a base $\mathscr{B}$ at $g$ and shown that $\langle\mathscr{B},\supseteq\rangle$ is a directed set, there’s only one reasonable thing to do. If $g$ really is in the closure of $E$, then it must be true that $B\cap E\ne\varnothing$ for each $B\in\mathscr{B}$, so for each $B\in\mathscr{B}$ we should be able to choose an $f_B\in B\cap E$.

Suppose that we can; then $\nu=\langle f_B:B\in\mathscr{B}\rangle$ is certainly a net in $E$. Does it converge to $g$? Let $U$ be any open nbhd of $g$; there is some $B_0\in\mathscr{B}$ such that $B_0\subseteq U$. If $B\in\mathscr{B}$, and $B_0\supseteq B$, then $$f_B\in B\cap E\subseteq B_0\subseteq U\;,$$ so the net $\nu$ is eventually in $U$: $f_B\in U$ for each $B\in\mathscr{B}$ with $B_0\supseteq B$. And $U$ was an arbitrary open nbhd of $g$, so $\nu$ converges to $g$, as desired. Thus, it only remains to show that $B\cap E\ne\varnothing$ for each $B\in\mathscr{B}$.

Assuming that you picked a natural local base at $g$, this is very easy. For instance, you might use the following local base. For each non-empty finite $F\subseteq\Bbb R$ and each real number $r>0$ let

$$B(F,r)=\left\{f\in\Bbb R^{\Bbb R}:|f(x)|<r\text{ for each }x\in F\right\}\;,$$

and let

$$\mathscr{B}=\left\{B(F,r):F\subseteq\Bbb R\text{ is finite and non-empty and }0<r\in\Bbb R\right\}\;.$$

It follows from the definition of the product topology that this $\mathscr{B}$ is a local base at $g$, and that automatically makes $\langle\mathscr{B},\supseteq\rangle$ a directed set: for any $B_0,B_1\in\mathscr{B}$ there is a $B_2\in\mathscr{B}$ such that $B_2\subseteq B_0\cap B_1$, i.e., such that $B_0\supseteq B_2$ and $B_1\supseteq B_2$, and $B_2$ is therefore an upper bound for $\{B_0,B_1\}$ in $\langle\mathscr{B},\supseteq\rangle$.

If you were feeling especially energetic, you could look a bit closer at this particular local base. Suppose that $B(F,r),B(G,s)\in\mathscr{B}$; then $$B(F,r)\cap B(G,s)\supseteq B(F\cup G,\min\{r,s\})\;,$$ so $B(F\cup G,\min\{r,s\})$ is an upper bound for $B(F,r)$ and $B(G,s)$. But this is unnecessary so long as you know that $\mathscr{B}$ is a base at $g$.

And now it’s easy to pick an $f_B\in B\cap E$ for each $B\in\mathscr{B}$: if $B=B(F,r)$, let

$$f_B:\Bbb R\to\Bbb R:x\mapsto\begin{cases} 0,&\text{if }x\in F\\ 1,&\text{if }x\in\Bbb R\setminus F\;. \end{cases}$$

Clearly $f_B\in B(F,r)\cap E$.

The only part of the argument that depends on the specific example is choosing the local base $\mathscr{B}$ at $g$ in such a way that we can easily find an $f_B\in B\cap E$ for each $B\in\mathscr{B}$; everything else falls out automatically from the fact that $\mathscr{B}$ is a local base at $g$.


Define a relation $\trianglelefteq$ on $E$ by putting $f_1\trianglelefteq f_2$ iff $f_1^{-1}(0)\subseteq f_2^{-1}(0)$, then by the construction of $E$ you have that $(E,\trianglelefteq)$ is a directed set.

Let $U$ be basic neighborhood of $g$ in $\Bbb R^{\Bbb R}$, then there are $a_1,\ldots,a_n\in \Bbb R$ and open sets $U_{a_1},\ldots, U_{a_n}$ of $\Bbb R$ with $U=\pi_{a_1}(U_{a_1})\cap \cdots \cap \pi_{a_n}(U_{a_n})$, then $0\in U_{a_i}$ for all $i$, let $f$ be given by $f(a_i)=0$ for all $i$ and $f(x)=1$ otherwise, then it is clear that for any $h\trianglerighteq f$ we have $h\in U$. So the net $(f)_{f\in (E;\trianglelefteq)}$ converges to $g$.

  • $\begingroup$ No, actually that was the last thing that I cannot see, I have something similar to what you said $\endgroup$ – Akaichan Apr 28 '13 at 23:03

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