Let $\text{Cont}(\mathbb{R},\mathbb{R})$ denote the set of continuous self-maps of $\mathbb{R}$ and let $\mathbb{R}^\mathbb{R}$ denote the set of all self-maps of $\mathbb{R}$, endowed with the product topology. Is $\text{Cont}(\mathbb{R},\mathbb{R})$ dense in $\mathbb{R}^\mathbb{R}$?
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I would say no, since a non Borel self-map of $\mathbb{R}$ is not the limit by pointwise convergence of continuous functions. – Phil-W Nov 23 '17 at 09:51
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If you can elaborate on this a bit, you can post it as an answer and we can close this thread. – Dominic van der Zypen Nov 23 '17 at 09:54
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7@Phil-W I think, this argument only shows that $\operatorname{Cont}(\mathbb{R},\mathbb{R})$ is not sequentially dense in $\mathbb{R}^{\mathbb{R}}$. – Jochen Glueck Nov 23 '17 at 10:05
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"this argument" is indeed not well-written: it should be "a non-Borel self-map of $\mathbf{R}$ is not the pointwise limit of a sequence of continuous functions". This example shows that one has to be careful with dealing with limits in non-metrizable spaces, such as, typically, uncountable products. – YCor Nov 23 '17 at 12:37
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1Additional exercise: there's a countable subset of $\mathrm{Cont}(\mathbf{R},\mathbf{R})$ that is dense in $\mathbf{R}^\mathbf{R}$. – YCor Nov 23 '17 at 13:06
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This question seems to be posed too quickly (without substantial preliminary thinking) and has an (almost trivial) affirmative answer: We should prove that $\mathrm{Cont}(\mathbb R,\mathbb R)$ intersects each non-empty open set $U\subset \mathbb R^{\mathbb R}$. We can assume that $U$ is of basic form: $U=\prod_{r\in\mathbb R}U_r$, where for each $r\in\mathbb R$ the set $U_r$ is open in $\mathbb R$ and the set $F=\{r\in\mathbb R:U_r\ne\mathbb R\}$ is finite. Now take any (piecewise linear) continuous function $f:\mathbb R\to\mathbb R$ such that $f(r)\in U_r$ for any $r\in F$. Then $f\in \mathrm{Cont}(\mathbb R,\mathbb R)\cap U$. So, $\mathrm{Cont}(\mathbb R,\mathbb R)$ is dense in $\mathbb R^{\mathbb R}$ (even for the Tychonoff product topology of the real lines endowed with the discrete topology).
Taras Banakh
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15$+1$ for your first sentence. This is the $593$rd question of Dominic Van der Zypen on MO, not counting the deleted ones... – js21 Nov 23 '17 at 10:14
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15On the other hand... MO would be a dreary site if it truly only contained research level questions with non trivial solutions. – Yaakov Baruch Nov 23 '17 at 10:32
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Thanks @YaakovBaruch, but I agree with Taras and js21 that I have asked this question without too much preliminary thinking – Dominic van der Zypen Nov 23 '17 at 10:35
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@Js21 By the way, according to the MO-profile of Dominic van der Zypen this is only 503rd his question (not 593rd). – Taras Banakh Nov 23 '17 at 12:38
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(@TarasBanakh 9 and 0 is close on the keyboard, but thanks for noticing!) – Dominic van der Zypen Nov 23 '17 at 12:52
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@DominicvanderZypen On the other hand, you have 503 questions and 90 answers, which sums up to the same 593 (and this cannot be explained by the nearness of 9 and 0 on the keyboard:) – Taras Banakh Nov 23 '17 at 12:56
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Yes.
Let $g: \mathbb{R} \to \mathbb{R}$ be an abritrary function.
Let $\mathcal{F}$ denote the set of all finite subsets of $\mathbb{R}$. We endow $\mathcal{F}$ with the order $\subseteq$, which renders it a directed set.
For each $F \in \mathcal{F}$, choose a continuous function $f_F$ which fulfils $f_F(x) = g(x)$ for all $x \in F$. Then the net $(f_F)_{F \in \mathcal{F}}$ converges pointwise to $g$.
Jochen Glueck
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