# Path connectedness of a particular algebraic set

Let $n\geq 3$, let $a$ and $b$ be two positive numbers, and

$$\Omega = \bigg\lbrace (x_1,x_2, \ldots ,x_n) \in ]0,+\infty[^n \ \bigg| \ x_1+x_2+ \ldots +x_n=a, \ x_1x_2 \ldots x_n=b \bigg\rbrace$$ It is easy to see that $\Omega$ is nonempty iff $\frac{a}{n} \geq b^{\frac{1}{n}}$. It seems clear enough that $\Omega$ is path connected, but I could find no simple proof of that.

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Assume $\frac{a}{n} \geq b^{1/n}$ so that $\Omega \neq \emptyset$.

Let $H$ be the hyperplane of $\mathbb{R}^n$ defined by the equation $x_1+x_2+\cdots +x_n = a$, and let $B = \{ (x_i) \in (0, \infty)^n \;|\; x_1x_2 \cdots x_n \geq b\}$. It should be clear that $\Omega = H \cap \partial B = \partial(H \cap B)$. Let's first consider the space $U = H \cap B$. You can show that $B$ is convex of dimension $n$, and since $H$ is a hyperplane, we have that $U$ is convex of dimension $\leq n-1$, hence path-connected. Furthermore, $U$ is bounded, since $H \cap B \subseteq (0,a)^n$

Now if $dim(U) < n-1$, then $U$ is a single point because $\partial B$ contains no line segment (see below for more details on this). In that case, $\Omega$ is definitely connected. On the other hand, when $dim(U) = n-1$, convexity and boundedness imply that $\partial U$ is homeomorphic to $S^{n-2}$. Since you have $n \geq 3$, we find that $\Omega = \partial U$ is connected.

Lemma: In $\mathbb{R}^n$, If a hyperplane $H$ intersects a convex set $B$ of dimension $n$ in such a way that there is an interior point $x$ of $B$ in the intersection, then $dim(H \cap B) = n - 1$.

Proof: If $x$ is interior to $B$, then there is an $n$-ball $V_x \subset B$ containing $x$. Suppose $V_x = \{ y \in \mathbb{R}^n \;|\; dist(x,y) \leq r\}$. Now $V_x \cap H = \{ y \in H \;|\; dist(x,y) \leq r\}$. Thus, $V_x \cap H$ is an $(n-1)$-dimensional ball in $H \cap B$.

Cor: If $dim(H \cap B) < n-1$, then $H \cap B \subseteq H \cap \partial B$.

Lemma: If $B \subseteq \mathbb{R}^n$ is convex and $H$ is a hyperplane such that $x \in H \cap B \subseteq H \cap \partial B$, and $\partial B$ contains no line segment, then $H \cap \partial B = \{ x\}$.

Proof: If $y \neq x$ is another point of $H \cap \partial B$, then we have two points $x, y \in H \cap B$, a convex set. Thus the line segment $\ell$ joining $x$ and $y$ is in $H \cap B \subseteq H \cap \partial B$. In particular, $\ell \in \partial B$, contradicting that $\partial B$ contains no segment.

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Nice solution! I’ll take some time to check the details before accepting it. –  Ewan Delanoy Dec 30 '12 at 17:59
"codimension 1 hyperplane" is redundant, right? –  Ewan Delanoy Dec 30 '12 at 17:59
To explain why $H$ is bounded, I would simply say that it is bounded because $H$ is (indeed, $H$ is included in the hypercube $[0,a]^n$). –  Ewan Delanoy Dec 30 '12 at 18:04
@EwanDelanoy you're correct on both of your suggestions... "hyperplane" is sufficient. And $H$ bounded is sufficient. –  Shaun Ault Dec 30 '12 at 18:27
Actually there are two different "H"s there : the hyperplane defined by the equation $x_1+x_2+ \ldots x_n=a$ and its intersection with $(0,\infty)^n$ (perhaps you could denote this by $H’$) –  Ewan Delanoy Dec 31 '12 at 6:15