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In $C[0,1]$ the set $\{f(x): f(0)\neq 0\}$ is dense? I know only that polynomials are dense in $C[0,1]$, could any one give me hint how to show this set is dense?thank you.

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it is dense${}$ – Norbert Aug 7 '12 at 19:20
For each $f \in C[0, 1]$, just observe that $f + \epsilon$ is in that set if $0 < |\epsilon|$ is sufficiently small. – Sangchul Lee Aug 7 '12 at 19:28
up vote 7 down vote accepted

Yes. Take $f\in \mathcal{C}[0,1]$ so that $f(0) = 0$. Now define $$f_n(x) = f(x) + {1\over n}, \qquad n\in\mathbb{N}.$$ We have $f_n\to f$ uniformly, whilst $f_n(0) \not= 0$ for all $n\in \mathbb{N}.$

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+1 Damn, that was beautiful, elegant and short...and even true. – DonAntonio Aug 7 '12 at 22:02

It is dense. It is enough to show that there are functions of arbitrarily small norm in the set, which shouldn't be too hard.

In general, the complement of any proper subspace of a normed space is dense, by a similar argument (possibly even easier, as is sometimes the case with generalizations where you don't think about the unnecessary things that would distract you otherwise).

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Your set $A = \{f\in C[0,1],f(0)\neq 0\}$ is dense in $C[0,1]$: let's take $f\in C[0,1]$ such as $f(0)=0$. We can consider the series $(f_n)$, where $$f_n(x) = \begin{cases}f(x) &\text{ if } x\geq\frac{1}{n} \\ \frac{1}{n}+n\times x\times \left(f\left(\frac{1}{n}\right)-\frac{1}{n}\right)& \text{ otherwise }\end{cases}$$ otherwise.

It is easy to verify that $\forall n, f_n\in A$, and $f_n\rightarrow f$ (for different distances), which proves that $A$ is dense in $C[0,1]$.

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I formatted your answer; please verify if my edits introduced an error. – M Turgeon Aug 7 '12 at 19:40
@MTurgeon yes, it's fine, it's much neater like that, thanks! – S4M Aug 7 '12 at 19:43

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