Let $H$ be a infinite dimensional, separable Hilbert space over $\mathbb{C}$


Let $B$ a subset of $H$ such that $B$ is linearly independent and such that exists a homeomorphism $f : [0,1] \to B$ between the unit interval $[0,1]$ and $B$


I would like to know if is it true that:


For every fixed $x \in (0,1)$
$$
\overline{
\operatorname{span}
f([0,x])
}
\cap
\overline{
\operatorname{span}
f((x,1])
}
=
\operatorname{span}
f(x)
$$



A homeomorphism between the unit interval $[0,1]$ and a linearly independent subset of a Hilbert space
It's not true.




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Edit: thinking twice there's a simpler solution.


On $L^2([0,1])$, define $f(x)(t)=1+t^{1/2x}$ for $x\in\mathopen]0,1]$ and $f(0)=1$. It's free (because the $t\mapsto t^x$, $x\ge 0$, form a free family). It depends continuously on $x$ (check at $x=0$). For $x\le 1/2$ it contains all (restrictions of) polynomials hence spans a dense subspace. So for $x=1/2$ the span intersection condition does not hold.




---


Previous solution (in the lines of my initial comment).


In $L^2([-1,1])$, define, for $0\le t\le 1/2$, $f(t)=\mathbb{1}\_{[-1,2t]}$.


It's clearly algebraically free. The closure of its span is $\mathbf{C}\mathbb{1}\_{[-1,0]}\oplus L^2([0,1])$ (indeed it contains, for $s<t$, $f(t/2)-f(s/2)=\mathbb{1}\_{[s,t]}$, etc).


Let $K$ be a Cantor subset in $[0,1]$ of measure $1/2$. There exists an increasing (hence measurable) map $g:[0,1]\to K$ such that for every $t$, the measure of $[0,g(t)]\cap K$ is equal to $t/2$. Define, for $t\in [1/2,1]$, $$f(t)=\mathbb{1}\_{[-1,1]}+\mathbb{1}\_{[0,g(2t-1)]\cap K}.$$
Then $f$ is continuous on $[0,1]$.


Clearly it does not satisfy the span intersection condition at $x=1/2$.


What remains is to show that $(f(t))\_{0\le t\le 1}$ is free. Consider a nontrivial linear relation: it can be written as
$$\sum\_{i=1}^na\_if(t\_i)=\sum\_{j=1}^mb\_jf(u\_j),\quad\text{ with }0\le t\_1<\dots<t\_n\le 1/2,\; 1/2<u\_1<\dots<u\_m\le 1,$$
with all $a\_i,b\_j$ nonzero and $\max(m,n)\ge 1$. 
Since $(f(t))\_{0\le t\le 1/2}$ is free, $m\ge 1$; write the linear dependence assumption as the equality in $L^2([-1,1])$:
$$\sum\_{i=1}^na\_if(t\_i)-\sum\_{j=1}^{m-1}b\_jf(u\_j)=b\_mf(u\_m).$$
The left-hand term is piecewise constant on $[g(u\_{m-1}),1]$ (or $[0,1]$ if $m=1$) while the right-hand term is not: if it were, $\mathbb{1}\_L$ would be (almost everywhere) equal to a piecewise constant function, where $L=[g(u\_{m-1}),1]\cap K$; since $L$ has positive mesure and its support has empty interior, this is not possible. 
This thus have a contradiction; accordingly $(f(t))\_{0\le t\le 1}$ is indeed free.


