# What is the basic nbhd of $C_p(X)$?

$C_p(X)$ denotes the set of all real-valued continuous functions on $X$ endowed with the topology of pointwise convergence. As the title explains, What is the basic nbhd of $C_p(X)$?

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A basic nbhd around a function $f$ is of the following form $$U_{x_1,\cdots, x_n, \varepsilon}=\{ g\in C_p(X); |f(x_i)-g(x_i)|<\varepsilon, \forall 1\leq i\leq n \},$$ where $x_1,\cdots, x_n$ are arbitrary points in $X$ and $\varepsilon>0$.

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Why? Could you say more? – Paul Feb 9 '13 at 2:29
We know that $f_n\rightarrow f$ in point wise topology if and only if $f_n(x)\rightarrow f(x)$ for all $x\in X$. This implies that the pointwise topology is equivalent to the topology induced by seminorms of the form $\rho_x: f\mapsto |f(x)|$ for $x\in X$. The above nbhd is a basic nbhd of the topology induced by this set of seminorms. – Vahid Shirbisheh Feb 9 '13 at 2:36
You can also use the fact that the topology of pointwise convergence is the one induced from the product topology on $\mathbb{R}^X$, and the basic open sets there are product sets that have only a non-trivial (not $\mathbb{R}$) open set on finitely many coordinates. The coordinates correspond to point evaluations and as basic open sets in $\mathbb{R}$ we can use symmetric open intervals around a point, etc. But @VahidShirbisheh argument is fine too. It depends on what you already know. – Henno Brandsma Feb 9 '13 at 6:01