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Artin, Algebra, Chapter 2, M6

It is intuitively easy but I am not sure if the writing my proof is rigorous enough.

Let $a=(a_1,a_2,\cdots,a_k)$ and $b=(b_1,b_2,\cdots,b_k)$ be points in $\mathbb{R}^k$. A path from $a$ to $b$ is a continuous function on the unit interval $[0,1] \to \mathbb{R}^k$. A function $X:[0,1]\to \mathbb{R}^k$ sending $t$ to $X(t)=(x_1(t),x_2(t),\cdots,x_k(t))$, such $X(0)=a,X(1)=b$. If $S$ is a subset of $\mathbb{R}^k$. We say that it is path-connected if any $a,b \in S$ can be joined by a path lying entirely in $S$.

Show that $\{xy=1\}$ is not path connected.

Proof:

Choose $a=(1,1), b=(-1,-1)$. Since $x_1(t)$ is continuous, there exists $0<c<1$ such that $x_1(c)=0$. But $(0,x_2(c))\notin S$ for all $x_2(c)$.

bof
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velut luna
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1 Answers1

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Your continuity argument is definitely correct. However, if you know some topology, there is even a slicker way to this end. Suppose the space $S$ were path-connected (I'll assume you are using usual Euclidean topology here), then it must also be connected, meaning that one could in no way find a partition $S=A\cup B$ in the sense that $A\cap \bar B=B\cap \bar A=\varnothing$, but this is apparently absurd since we can immediately find out that $S$ consists of two disjoint branches which form such a partition, meaning that $S$ is indeed a non-connected space.

Vim
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