some confusion regarding vector spaces..

Question

let $C[0,1]$ with supremum norm then , $$S =\left\{f ∈ C[0,1] :\lim \limits_{n\to \infty} f(1/n) = 0\right\}$$

then which of the following is correct

a) vector space but not closed in $C[0,1]$

b) closed but doesnot form a vector space

c) a closed vector sapce but not a algebra

d) a closed algebra

i think option 1 is correct as it is vector space but not closed

Any hints or ideas or solution will apprecaited

thanks in advance

Answer 1

It is a closed subalgebra.

Suppose that $f\in C\bigl([0,1]\bigr)\setminus S$. Then $\lim_{n\to\infty}f(1/n)\neq0$. So, you can take some $\varepsilon>0$ such that, for each $p\in\mathbb N$, there is some $n\geqslant p$ such that $\bigl|f(1/n)\bigr|\geqslant\varepsilon$. But then every $g\in B(f,\varepsilon/2)$ also belongs to $C\bigl([0,1]\bigr)\setminus S$. So, this set is open and therefore $S$ is closed.

It should be clear that if $f,g\in S$ and $\alpha,\beta\in\mathbb R$, then $\alpha f+\beta g$ and $fg$ also belongs to $S$. Therefore, $S$ is a subalgebra (with respect to the usual product of functions).

Answer 2

Assume that $f_{m}\rightarrow f$ in that norm, then given $\epsilon>0$, we have $\|f-f_{m}\|<\epsilon$ for some $m$. So $|f(1/n)-f_{m}(1/n)|<\epsilon$, taking $n\rightarrow\infty$ we conclude that $f(1/n)\rightarrow 0$, so $f\in S$, $S$ is closed.