I know that the set of algebraic polynomials is dense in space $L_{p}(a,b)$, where $1 \leq p < \infty$ and $a,b \in \mathbb{R}$. However, what about $L_{\infty}(0,1)$? Will the set of algebraic polynomials be dense in space $L_{\infty}(0,1)$?

In my opinion it is not, but is there any example that prove this fact?

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    $\begingroup$ Since $L_{\infty}(0,1)$ is not separable, it cannot have the set of real polynomials as a dense subset. If it were true, the set of algebraic polynomials with rational coefficients would be dense in $L_{\infty}(0,1)$. But that is not possible since the set of rational polynomials is countable. $\endgroup$ – Darth Lubinus Jun 3 at 18:48
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    $\begingroup$ @DarthLubinus, Could you explain me why this is impossible? $\endgroup$ – Sh VavilenT Jun 3 at 19:52
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    $\begingroup$ Sure, I'll give some details in an answer! $\endgroup$ – Darth Lubinus Jun 3 at 19:55

I'm going to try and ellaborate on my previous comment.

First, let's call $\mathbb{R}[X]$ the ring of real polynomials, and suppose $\mathbb{R}[X]$ was a dense subset of $L_{\infty}(0,1)$. Therefore, we can prove that $\mathbb{Q}[X]$ (the subset of all rational polynomials) is dense in $L_{\infty}(0,1)$ by seeing that it is dense in $\mathbb{R}[X]$.

Let $p(x)=a_{n}x^{n}+...+a_{1}x+a_{0}$ be a polynomial with coefficients in $\mathbb{R}$, and $\varepsilon>0$. Then, for every $a_{i}$ we can find a $b_{i}\in \mathbb{Q}$ such that $|b_{i}-a_{i}|<\varepsilon/n$. By taking $q(x)=b_{n}x^{n}+...+b_{1}x+b_{0}$, we have that $q\in \mathbb{Q}[X]$, and for every $x\in [0,1]$,

$$|p(x)-q(x)|=\Big|\sum_{i=0}^{n}(a_{i}-b_{i})x\Big|\leq \sum_{i=0}^{n}|a_{i}-b_{i}||x|\leq n\frac{\varepsilon}{n}=\varepsilon,$$

so $||p-q||_{\infty}\leq\varepsilon$. Therefore, $\mathbb{Q}[X]$ is dense in $\mathbb{R}[X]$ and, using our assumption that $\mathbb{R}[X]$ is dense in $L_{\infty}(0,1)$, we conclude that $\mathbb{Q}[X]$ is dense in $L_{\infty}(0,1)$. In particular, $L_{\infty}(0,1)$ admits a countable dense subspace, which means that it is separable.

To find our desired contradiction, we need to prove that $L_{\infty}(0,1)$ cannot be separable.

Consider the family $\{f_{t}\colon t\in (0,1)\}\subseteq L_{\infty}(0,1)$, where $f_{t}(s)=1$ if $s\leq t$, and $f_{t}(s)=0$ otherwise. It is easy to see that, whenever $t\neq s$, we have $||f_{t}-f_{s}||_{\infty}=1$. Therefore, we can find an uncountable family of disjoint open balls $A_{t}=B(f_{t},1/3)$, where $0 < t < 1$. Since $\mathbb{Q}[X]$ is dense in $L_{\infty}(0,1)$, each $A_{t}$ must contain at least one polynomial $q_{t}\in \mathbb{Q}[X]$, but this is impossible, because it would imply that there is an uncountable amount of rational polynomials! Therefore, $\mathbb{R}[X]$ cannot be dense.

In essence, what is happening is that $L_{\infty}$ is too big of a space to be the closure of something as comparably small as a countable subset. When $p<\infty$, this doesn't happen, and the fact that polynomials are dense in $L_{p}(0,1)$ proves that this last space is separable.

Just for the sake of adding another point of view, here is a more "topological" proof (not too different from the one I just did). In metric spaces, separability is the same thing as second countability (the property of having a countable basis for the topology of the metric space). If $L_{\infty}(0,1)$ were separable/secound countable, then the family $B=\{f_{t}\}_{t\in (0,1)}$ would be another second countable space (as a subspace of one). But $||f_{t}-f_{s}||_{\infty}=1$ when $t\neq s$, so $B$ must be discrete. In that case, $B$ must be countable, which is again a contradiction.

I hope this helps!

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