samples/texts/102047/page_16.md

with $\operatorname{span} X' = \mathbb{R}^s$, and consider the map

$$T:I^{X^{\mathrm{\prime }}}\to \prod _{h\in \mathbb{H}(X^{\mathrm{\prime }})}{P}_h:q\mapsto (p_h(D)q{)}_{h\in \mathbb{H}(X^{\mathrm{\prime }})},T:\; I^\{X\text{'}\}\; \backslash to\; \backslash prod\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\}\; P\_h\; :\; q\; \backslash mapsto\; (p\_h(D)q)\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\},$$

where $p_h := p_{h, X'} = \langle h^\perp, \cdot \rangle^{(X'\setminus h)}$ are the generators of $I^{X'}$ and $P_h := p_h(D)(I^{X'}\perp)$. Then

$$\dim I^{X^{\mathrm{\prime }}}\le \dim I^{X^{\mathrm{\prime }}}\perp +\sum _{h\in \mathbb{H}(X^{\mathrm{\prime }})}\dim P_h,\backslash dim\; I^\{X\text{'}\}\; \backslash leq\; \backslash dim\; I^\{X\text{'}\}\; \backslash perp\; +\; \backslash sum\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\}\; \backslash dim\; P\_h,$$

since

$$\ker T=(I^{X^{\mathrm{\prime }}}\perp )\cap (I^X\perp )\subseteq I^{X^{\mathrm{\prime }}}\perp \mathrm{.}\backslash ker\; T\; =\; (I^\{X\text{'}\}\; \backslash perp)\; \backslash cap\; (I^\{X\}\; \backslash perp)\; \backslash subseteq\; I^\{X\text{'}\}\; \backslash perp.$$

Consequently, by induction,

$$\dim (I^{X^{\mathrm{\prime }}}\perp )\le b(X^{\mathrm{\prime }})+\sum _{h\in \mathbb{H}(X^{\mathrm{\prime }})}\dim P_h\mathrm{.}\backslash dim(I^\{X\text{'}\}\; \backslash perp)\; \backslash leq\; b(X\text{'})\; +\; \backslash sum\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\}\; \backslash dim\; P\_h.$$

This proves that

$$\dim I^{X^{\mathrm{\prime }}}\perp \le b(X),\backslash dim\; I^\{X\text{'}\}\; \backslash perp\; \backslash leq\; b(X),$$

provided we can prove that

$$\sum _{h\in \mathbb{H}(X^{\mathrm{\prime }})}\dim P_h\le \mathrm{\#}\{B\in \mathcal{B}(X):x\in B\}\mathrm{.}\backslash sum\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\}\; \backslash dim\; P\_h\; \backslash leq\; \backslash \#\backslash \{B\; \backslash in\; \backslash mathcal\{B\}(X)\; :\; x\; \backslash in\; B\backslash \}.$$

In particular, it is sufficient to prove that, for all $h \in \mathbb{H}(X')$,

$$(2.12)P_h\subset I^{X_h}\perp (2.12)\; \backslash qquad\; P\_h\; \backslash subset\; I^\{X\_h\}\; \backslash perp$$

with

$$X_h:=(X\cap h)\cup x\mathrm{.}X\_h\; :=\; (X\; \backslash cap\; h)\; \backslash cup\; x.$$

For, (2.12) implies that

$$\dim P_h\le \dim I^{X_h}\perp \le b(X_h),\backslash dim\; P\_h\; \backslash leq\; \backslash dim\; I^\{X\_h\}\; \backslash perp\; \backslash leq\; b(X\_h),$$

(the last inequality by induction), while

$$\sum _{h\in \mathbb{H}(X^{\mathrm{\prime }})}b(X_h)=\mathrm{\#}\{B\in \mathcal{B}(X):x\in B\}\mathrm{.}\backslash sum\_\{h\; \backslash in\; \backslash mathbb\{H\}(X\text{'})\}\; b(X\_h)\; =\; \backslash \#\backslash \{B\; \backslash in\; \backslash mathcal\{B\}(X)\; :\; x\; \backslash in\; B\backslash \}.$$

The claim (2.12) is trivial in case $x \in h$, since then $X'\setminus h = X\setminus h$, and therefore $p_{h,X'} = p_{h,X}$ and so $P_h = {0}$ in that case.

We now prove (2.12) for the contrary case, i.e., the case when $x \notin h$. We have to show that for every $k \in \mathbb{H}(X_h)$

$$(2.13)p_{k,X_h}(D)P_h=\{0\},(2.13)\; \backslash qquad\; p\_\{k,\; X\_h\}(D)P\_h\; =\; \backslash \{0\backslash \},$$