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<div id="p1" class="ltx_para">
<p class="ltx_p">This proof uses the Borsuk-Ulam theorem, which states that any continuous function from <math id="p1.m1" class="ltx_Math" alttext="S^{n}" display="inline"><msup><mi>S</mi><mi>n</mi></msup></math> to <math id="p1.m2" class="ltx_Math" alttext="\mathbb{R}^{n}" display="inline"><msup><mi>ℝ</mi><mi>n</mi></msup></math> maps some pair of antipodal points to the same point.</p>
</div>
<div id="p2" class="ltx_para">
<p class="ltx_p">Let <math id="p2.m1" class="ltx_Math" alttext="A" display="inline"><mi>A</mi></math> be a measurable bounded subset of <math id="p2.m2" class="ltx_Math" alttext="\mathbb{R}^{n}" display="inline"><msup><mi>ℝ</mi><mi>n</mi></msup></math>. Given any unit vector <math id="p2.m3" class="ltx_Math" alttext="\hat{n}\in S^{{n-1}}" display="inline"><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>∈</mo><msup><mi>S</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup></mrow></math> and <math id="p2.m4" class="ltx_Math" alttext="s\in\mathbb{R}" display="inline"><mrow><mi>s</mi><mo>∈</mo><mi>ℝ</mi></mrow></math>, there is a unique <math id="p2.m5" class="ltx_Math" alttext="n-1" display="inline"><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></math> dimensional hyperplane normal to <math id="p2.m6" class="ltx_Math" alttext="\hat{n}" display="inline"><mover accent="true"><mi>n</mi><mo>^</mo></mover></math> and containing <math id="p2.m7" class="ltx_Math" alttext="s\hat{n}" display="inline"><mrow><mi>s</mi><mo></mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow></math>.</p>
</div>
<div id="p3" class="ltx_para">
<p class="ltx_p">Define <math id="p3.m1" class="ltx_Math" alttext="f:S^{{n-1}}\times\mathbb{R}\rightarrow[0,\infty)" display="inline"><mrow><mi>f</mi><mo>:</mo><mrow><mrow><msup><mi>S</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup><mo>×</mo><mi>ℝ</mi></mrow><mo>→</mo><mrow><mo>[</mo><mrow><mn>0</mn><mo>,</mo><mi mathvariant="normal">∞</mi></mrow><mo>)</mo></mrow></mrow></mrow></math> by sending <math id="p3.m2" class="ltx_Math" alttext="(\hat{n},s)" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></math> to the measure of the subset of <math id="p3.m3" class="ltx_Math" alttext="A" display="inline"><mi>A</mi></math> lying on the side of the plane corresponding to <math id="p3.m4" class="ltx_Math" alttext="(\hat{n},s)" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></math> in the direction in which <math id="p3.m5" class="ltx_Math" alttext="\hat{n}" display="inline"><mover accent="true"><mi>n</mi><mo>^</mo></mover></math> points. Note that <math id="p3.m6" class="ltx_Math" alttext="(\hat{n},s)" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></math> and <math id="p3.m7" class="ltx_Math" alttext="(-\hat{n},-s)" display="inline"><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><mo>-</mo><mi>s</mi></mrow></mrow><mo>)</mo></mrow></math> correspond to the same plane, but to different sides of the plane, so that <math id="p3.m8" class="ltx_Math" alttext="f(\hat{n},s)+f(-\hat{n},-s)=m(A)" display="inline"><mrow><mrow><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow><mo>+</mo><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><mo>-</mo><mi>s</mi></mrow></mrow><mo>)</mo></mrow></mrow></mrow><mo>=</mo><mrow><mi>m</mi><mo></mo><mrow><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mrow></mrow></math>.</p>
</div>
<div id="p4" class="ltx_para">
<p class="ltx_p">Since <math id="p4.m1" class="ltx_Math" alttext="A" display="inline"><mi>A</mi></math> is bounded, there is an <math id="p4.m2" class="ltx_Math" alttext="r>0" display="inline"><mrow><mi>r</mi><mo>></mo><mn>0</mn></mrow></math> such that <math id="p4.m3" class="ltx_Math" alttext="A" display="inline"><mi>A</mi></math> is contained in <math id="p4.m4" class="ltx_Math" alttext="\overline{B_{r}}" display="inline"><mover accent="true"><msub><mi>B</mi><mi>r</mi></msub><mo>¯</mo></mover></math>, the closed ball of radius <math id="p4.m5" class="ltx_Math" alttext="r" display="inline"><mi>r</mi></math> centered at the origin. For sufficiently small changes in <math id="p4.m6" class="ltx_Math" alttext="(\hat{n},s)" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></math>, the measure of the portion of <math id="p4.m7" class="ltx_Math" alttext="\overline{B_{r}}" display="inline"><mover accent="true"><msub><mi>B</mi><mi>r</mi></msub><mo>¯</mo></mover></math> between the different corresponding planes can be made arbitrarily small, and this bounds the change in <math id="p4.m8" class="ltx_Math" alttext="f(\hat{n},s)" display="inline"><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow></math>, so that <math id="p4.m9" class="ltx_Math" alttext="f" display="inline"><mi>f</mi></math> is a continuous function.</p>
</div>
<div id="p5" class="ltx_para">
<p class="ltx_p">Finally, it’s easy to see that, for fixed <math id="p5.m1" class="ltx_Math" alttext="\hat{n}" display="inline"><mover accent="true"><mi>n</mi><mo>^</mo></mover></math>, <math id="p5.m2" class="ltx_Math" alttext="f(\hat{n},s)" display="inline"><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow></math> is monotonically decreasing in <math id="p5.m3" class="ltx_Math" alttext="s" display="inline"><mi>s</mi></math>, with <math id="p5.m4" class="ltx_Math" alttext="f(\hat{n},-s)=m(A)" display="inline"><mrow><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><mo>-</mo><mi>s</mi></mrow></mrow><mo>)</mo></mrow></mrow><mo>=</mo><mrow><mi>m</mi><mo></mo><mrow><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mrow></mrow></math> and <math id="p5.m5" class="ltx_Math" alttext="f(\hat{n},s)=0" display="inline"><mrow><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow><mo>=</mo><mn>0</mn></mrow></math> for <math id="p5.m6" class="ltx_Math" alttext="s" display="inline"><mi>s</mi></math> sufficiently large.</p>
</div>
<div id="p6" class="ltx_para">
<p class="ltx_p">Given these properties of <math id="p6.m1" class="ltx_Math" alttext="f" display="inline"><mi>f</mi></math>, we see by the intermediate value theorem that, for fixed <math id="p6.m2" class="ltx_Math" alttext="\hat{n}" display="inline"><mover accent="true"><mi>n</mi><mo>^</mo></mover></math>, there is an interval <math id="p6.m3" class="ltx_Math" alttext="[a,b]" display="inline"><mrow><mo>[</mo><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow><mo>]</mo></mrow></math> such that the set of <math id="p6.m4" class="ltx_Math" alttext="s" display="inline"><mi>s</mi></math> with <math id="p6.m5" class="ltx_Math" alttext="f(\hat{n},s)=m(A)/2" display="inline"><mrow><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow><mo>=</mo><mrow><mrow><mi>m</mi><mo></mo><mrow><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mrow><mo>/</mo><mn>2</mn></mrow></mrow></math> is <math id="p6.m6" class="ltx_Math" alttext="[a,b]" display="inline"><mrow><mo>[</mo><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow><mo>]</mo></mrow></math>. If we define <math id="p6.m7" class="ltx_Math" alttext="g(\hat{n})" display="inline"><mrow><mi>g</mi><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></math> to be the midpoint of this interval, then, since <math id="p6.m8" class="ltx_Math" alttext="f" display="inline"><mi>f</mi></math> is continuous, we see <math id="p6.m9" class="ltx_Math" alttext="g" display="inline"><mi>g</mi></math> is a continuous function from <math id="p6.m10" class="ltx_Math" alttext="S^{{n-1}}" display="inline"><msup><mi>S</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup></math> to <math id="p6.m11" class="ltx_Math" alttext="\mathbb{R}" display="inline"><mi>ℝ</mi></math>. Also, since <math id="p6.m12" class="ltx_Math" alttext="f(\hat{n},s)+f(-\hat{n},-s)=m(A)" display="inline"><mrow><mrow><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mi>s</mi></mrow><mo>)</mo></mrow></mrow><mo>+</mo><mrow><mi>f</mi><mo></mo><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><mo>-</mo><mi>s</mi></mrow></mrow><mo>)</mo></mrow></mrow></mrow><mo>=</mo><mrow><mi>m</mi><mo></mo><mrow><mo>(</mo><mi>A</mi><mo>)</mo></mrow></mrow></mrow></math>, if <math id="p6.m13" class="ltx_Math" alttext="[a,b]" display="inline"><mrow><mo>[</mo><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow><mo>]</mo></mrow></math> is the interval corresponding to <math id="p6.m14" class="ltx_Math" alttext="\hat{n}" display="inline"><mover accent="true"><mi>n</mi><mo>^</mo></mover></math>, then <math id="p6.m15" class="ltx_Math" alttext="[-b,-a]" display="inline"><mrow><mo>[</mo><mrow><mrow><mo>-</mo><mi>b</mi></mrow><mo>,</mo><mrow><mo>-</mo><mi>a</mi></mrow></mrow><mo>]</mo></mrow></math> is the interval corresponding to <math id="p6.m16" class="ltx_Math" alttext="-\hat{n}" display="inline"><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow></math>, and so <math id="p6.m17" class="ltx_Math" alttext="g(\hat{n})=-g(-\hat{n})" display="inline"><mrow><mrow><mi>g</mi><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow><mo>=</mo><mrow><mo>-</mo><mrow><mi>g</mi><mo></mo><mrow><mo>(</mo><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>)</mo></mrow></mrow></mrow></mrow></math>.</p>
</div>
<div id="p7" class="ltx_para">
<p class="ltx_p">Now let <math id="p7.m1" class="ltx_Math" alttext="A_{1},A_{2},...,A_{n}" display="inline"><mrow><msub><mi>A</mi><mn>1</mn></msub><mo>,</mo><msub><mi>A</mi><mn>2</mn></msub><mo>,</mo><mi mathvariant="normal">…</mi><mo>,</mo><msub><mi>A</mi><mi>n</mi></msub></mrow></math> be measurable bounded subsets of <math id="p7.m2" class="ltx_Math" alttext="\mathbb{R}^{n}" display="inline"><msup><mi>ℝ</mi><mi>n</mi></msup></math>, and let <math id="p7.m3" class="ltx_Math" alttext="f_{i},g_{i}" display="inline"><mrow><msub><mi>f</mi><mi>i</mi></msub><mo>,</mo><msub><mi>g</mi><mi>i</mi></msub></mrow></math> be the maps constructed above for <math id="p7.m4" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math>. Then we can define <math id="p7.m5" class="ltx_Math" alttext="h:S^{{n-1}}\rightarrow R^{{n-1}}" display="inline"><mrow><mi>h</mi><mo>:</mo><mrow><msup><mi>S</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup><mo>→</mo><msup><mi>R</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup></mrow></mrow></math> by:</p>
</div>
<div id="p8" class="ltx_para">
<table id="S0.Ex1" class="ltx_equation">
<tr class="ltx_equation ltx_align_baseline">
<td class="ltx_eqn_pad"></td>
<td class="ltx_align_center"><math id="S0.Ex1.m1" class="ltx_Math" alttext="h(\hat{n})=(f_{1}(\hat{n},g_{n}(\hat{n})),f_{2}(\hat{n},g_{n}(\hat{n})),...f_{%
{n-1}}(\hat{n},g_{n}(\hat{n})))" display="block"><mrow><mrow><mi>h</mi><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow><mo>=</mo><mrow><mo>(</mo><mrow><mrow><msub><mi>f</mi><mn>1</mn></msub><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow><mo>,</mo><mrow><msub><mi>f</mi><mn>2</mn></msub><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow><mo>,</mo><mrow><mi mathvariant="normal">…</mi><mo></mo><msub><mi>f</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msub><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow></math></td>
<td class="ltx_eqn_pad"></td>
</tr>
</table>
</div>
<div id="p9" class="ltx_para">
<p class="ltx_p">This is continuous, since each coordinate function is the composition of continuous functions. Thus we can apply the Borsuk-Ulam theorem to see there is some <math id="p9.m1" class="ltx_Math" alttext="\hat{n}\in S^{{n-1}}" display="inline"><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>∈</mo><msup><mi>S</mi><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></msup></mrow></math> with <math id="p9.m2" class="ltx_Math" alttext="h(\hat{n})=h(-\hat{n})" display="inline"><mrow><mrow><mi>h</mi><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow><mo>=</mo><mrow><mi>h</mi><mo></mo><mrow><mo>(</mo><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>)</mo></mrow></mrow></mrow></math>, ie, with:</p>
</div>
<div id="p10" class="ltx_para">
<table id="S0.Ex2" class="ltx_equation">
<tr class="ltx_equation ltx_align_baseline">
<td class="ltx_eqn_pad"></td>
<td class="ltx_align_center"><math id="S0.Ex2.m1" class="ltx_Math" alttext="f_{i}(\hat{n},g_{n}(\hat{n}))=f_{i}(-\hat{n},g_{n}(-\hat{n}))=f_{i}(-\hat{n},-%
g_{n}(\hat{n}))" display="block"><mrow><mrow><msub><mi>f</mi><mi>i</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow><mo>=</mo><mrow><msub><mi>f</mi><mi>i</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow><mo>=</mo><mrow><msub><mi>f</mi><mi>i</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><mo>-</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow></mrow><mo>)</mo></mrow></mrow></mrow></math></td>
<td class="ltx_eqn_pad"></td>
</tr>
</table>
</div>
<div id="p11" class="ltx_para">
<p class="ltx_p">where we’ve used the property of <math id="p11.m1" class="ltx_Math" alttext="g" display="inline"><mi>g</mi></math> mentioned above. But this just means that for each <math id="p11.m2" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math> with <math id="p11.m3" class="ltx_Math" alttext="1\leq i\leq n-1" display="inline"><mrow><mn>1</mn><mo>≤</mo><mi>i</mi><mo>≤</mo><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></mrow></math>, the measure of the subset of <math id="p11.m4" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math> lying on one side of the plane corresponding to <math id="p11.m5" class="ltx_Math" alttext="(\hat{n},g_{n}(\hat{n}))" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></math>, which is <math id="p11.m6" class="ltx_Math" alttext="f_{i}(\hat{n},g_{n}(\hat{n}))" display="inline"><mrow><msub><mi>f</mi><mi>i</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></mrow></math>, is the same as the measure of the subset of <math id="p11.m7" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math> lying on the other side of the plane, which is <math id="p11.m8" class="ltx_Math" alttext="f_{i}(-\hat{n},-g_{n}(\hat{n}))" display="inline"><mrow><msub><mi>f</mi><mi>i</mi></msub><mo></mo><mrow><mo>(</mo><mrow><mrow><mo>-</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover></mrow><mo>,</mo><mrow><mo>-</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow></mrow><mo>)</mo></mrow></mrow></math>. In other words, the plane corresponding to <math id="p11.m9" class="ltx_Math" alttext="(\hat{n},g_{n}(\hat{n}))" display="inline"><mrow><mo>(</mo><mrow><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>,</mo><mrow><msub><mi>g</mi><mi>n</mi></msub><mo></mo><mrow><mo>(</mo><mover accent="true"><mi>n</mi><mo>^</mo></mover><mo>)</mo></mrow></mrow></mrow><mo>)</mo></mrow></math> bisects each <math id="p11.m10" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math> with <math id="p11.m11" class="ltx_Math" alttext="1\leq i\leq n-1" display="inline"><mrow><mn>1</mn><mo>≤</mo><mi>i</mi><mo>≤</mo><mrow><mi>n</mi><mo>-</mo><mn>1</mn></mrow></mrow></math>. Finally, by the definition of <math id="p11.m12" class="ltx_Math" alttext="g_{n}" display="inline"><msub><mi>g</mi><mi>n</mi></msub></math>, this plane also bisects <math id="p11.m13" class="ltx_Math" alttext="A_{n}" display="inline"><msub><mi>A</mi><mi>n</mi></msub></math>, and so it bisects each of the <math id="p11.m14" class="ltx_Math" alttext="A_{i}" display="inline"><msub><mi>A</mi><mi>i</mi></msub></math> as claimed.
</p>
</div>
</section>
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