Let $\mathcal H$ be a Hilbert space over $\mathbb{C}$. Suppose $A\subseteq B\subseteq\mathcal H$ and $A$ is dense in $B$. Then $span\,A$ is dense in $span\,B$(?).
I know that if $f:X\to Y$ is a continuous function and $A$ is dense in $X$, then $f(A)$ is dense in $f(X)$. So my attempt was to prove "$\operatorname{span}$" is a continuous function. But I'm kind of stuck here. I'm aware that addition and scalar multiplication is continuous, but since span consists of arbitrary (but finite) number of linear combinations, I don't have a clear idea on how to prove.
Any hints or counterexamples?
(Sorry if this is a silly question. I don't have much background in Functional Analysis.)
Take $v\in\operatorname{span}(B)\setminus\{0\}$ and take $\varepsilon>0$. There is a natural $n$ and there are vectors $v_1,\ldots,v_n\in B$ and non-zero scalars $\lambda_1,\ldots,\lambda_n$ such that$$\left\lVert v-\sum_{k=1}^n\lambda_kv_k\right\rVert<\frac\varepsilon2.$$For each $k\in\{1,2,\ldots,n\}$, there is some vector $w_k\in A$ such that $\lVert w_k-v_k\rVert<\frac\varepsilon{2n\lvert\lambda_k\rvert}$. But then$$\left\lVert\sum_{k=1}^n\lambda_kv_k-\sum_{k=1}^n\lambda_kw_k\right\rVert<\frac\varepsilon2$$and therefore$$\left\lVert v-\sum_{k=1}^n\lambda_kw_k\right\rVert<\varepsilon.$$And, of course, $\sum_{k=1}^n\lambda_kw_k\in\operatorname{span}(A)$.