Intersection of 'spheres' in Hilbert space with respect to real analytically moving mid points

Question

The intersection of (countably many) 'spheres' in a Hilbert space can be non-empty. If we make this situation moving real analytically, the mid points and the radii, can it happen that the intersection becomes empty while it is not empty on an open set of the parameter space?

Precisely:

Let $H$ be a (separable) complex Hilbert space and $ z_i:(0,2)\to H$ and $r_i:(0,2)\to (\delta,1)$ be real analytic maps for $i=1,2,...$, where $\delta>0$. Furthermore $z_i(t)\not=z_j(t)$ for all $t$ and $i\not =j$. The set $M_t=\{w\in H\ \vert\ \langle w,z_i(t) \rangle =r_i(t)\ \forall i\}$ is assumed to be not empty for $t\in (0,1)$. Is it possible that $M_1$ is empty?

Remark: I am aware that the sets I call 'spheres' in fact are not spheres. I couldn't think of a better word.

Thanks for any hint!

Answer 1

This is my new answer for the edited question

Here is a counterexample. Let $H = \mathbb{C}^2$ and let $z_1(t) = (1,t)$ and $z_i(t) = (2,2)$ for all $i > 1$ and for all $t$. Also let $r_i(t) = 1$ for all $i$ and $t$.

Then $M_t$ is the set of points $(z,w) \in \mathbb{C}^2$ such that $z + tw = 1$ and $2z + 2w = 1$. As long as $t \neq 1$, there is a unique point $(z,w)$ which satisfies this system of linear equations (i.e. there is one point in $M_t$, so it is not empty). However, when $t = 1$ the system becomes $$ z + w = 1 = 2 (z + w), $$ which clearly has no solutions, so $M_1$ is empty.

For what it's worth, the 'spheres' you talk of are simply hyperplanes in Hilbert space (affine spaces of codimension 1).

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This is my old answer to this question, before it was edited

If I'm interpreting your question right, then yes it is possible for $M_1$ to be empty (and in fact it is possible for $M_t$ to be empty for all $t \in (0,1)$ as well).

For example, consider $H = \mathbb{C}$, $z_i(t) = i$ and $r_i(t) = c$ for all $i$ and $t$, where $c$ is some number in $(\delta, 1)$. Then $M_t$ is the set of points $w \in \mathbb{C}$ such that $w i = c$ for all $i = 1, 2, \dotsc$, and of course there are no such $w$.