Let $C[0,1]$ be the space of all continuous functions on the interval $[0,1]$, then you can induce a topology of pointwise convergence, right? First question, how will these open sets look like? My other question, is this space metrizable? If not, why?
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The seminorms inducing the topology are given by $p_t(f) := |f(t)|$, where $t\in [0,1]$. Now, you can buid up your topology by defining the subbasis and then the basis. – Friedrich Philipp Mar 06 '16 at 14:45
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Is it possible to define i topology not induced by any norm in this space? – Amanda Mar 06 '16 at 14:48
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I don't understand your question. – Friedrich Philipp Mar 06 '16 at 14:53
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Welcome to Math.SE. Your opening sentence suggests the question, is a topology determined by its convergent sequences? Is it possible that two different topologies have the same convergent sequences? As the rest of your Question is related to metrizability, let's note that a metric topology is determined by its convergent sequences, but more generally sequences are not "enough" to do this. A related post at MathOverflow: Is a topology determined by its convergent sequences? – hardmath Mar 06 '16 at 14:57
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Possible duplicate of Deduce that the product of uncountably many copies of the real line \Bbb{R} is not metrizable. – Xiang Yu Mar 06 '16 at 15:37
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@XiangYu $C[0,1]$ is a non-closed proper subspace of $\mathbb R^{[0,1]}$. – Yai0Phah Mar 06 '16 at 20:25
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@FrankScience Sorry, I don't see the question clearly. – Xiang Yu Mar 07 '16 at 02:07
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Here are some useful results that answer this question and more general questions of the same nature.
Denote by $C_{p}(X)$ the collection of continuous functions $X\to\mathbb{R}$ with the topology of point wise convergence; define the weight $w(Y)$ of a topological space $Y$ to be the minimum cardinality of a basis; and define the character $\chi(Y):=\sup_{y}\chi(Y,y)$ of a topological space $Y$ to be the supremum over all minimal cardinalities of the neighborhood basis of $Y$, i.e. $\chi(Y,y)$ is the minimum cardinality of a neighborhood basis at $y\in Y$.
Let $Y$ then be an infinite set.
Theorem #1: we have $|Y|=\chi(C_{p}(Y))=w(C_{p}(Y))$.
Theorem #2: the following are equivalent for any infinite cardinal $\kappa$: \begin{align*} (i)& \;\; w(Y)\leq \kappa,\\ (ii)&\;\; Y\;\mathrm{embeds}\;\mathrm{in}\;\mathbb{R}^{\kappa},\\ (iii)&\;\; Y\;\mathrm{embeds}\;\mathrm{in}\;[0,1]^{\kappa}. \end{align*}
Now every metrizable space is first countable, so if $C_{p}(X)$ is metrizable we have by Theorem #1 that $|X|\leq \omega$. And conversely, if $X$ is countable, then by Theorem #1 we have $w(C_{p}(X))=\omega$, and Theorem #2 gives us an embedding $C_{p}(X)\to\mathbb{R}^{\omega}$, making $C_{p}(X)$ metrizable.
Conclusion: $C_{p}(X)$ is metrizable if and only if $X$ countable.
To your specific example, since $[0,1]$ is uncountable, then $C[0,1]$ with point wise convergence is not metrizable.
Reference to Theorems #1 and #2 is e.g. the following book: "V. Tkachuk - $C_{p}$ theory problem book of topological and function spaces", s.209 (p. 165) and s.169 (p. 142).
T. Eskin
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The topology of point-wise convergence in $C[0,1]$ is not metrisable because it is not first-countable. As for the open sets, this topology has a natural subbase consisting of sets of the form $$\{f\in C[0,1]\colon |f(x)|<\varepsilon\}\quad (x\in [0,1], \varepsilon > 0).$$
Tomasz Kania
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