Error on proof of :$\mathbb{R}^{n}$ can't be covered by open disjoint balls assuming euclidean norm

Question

I'm trying to prove the above afirmation and so far this was what i got:

Suppose it could be covered,then $R^n=\cup A_x$ with $x\in X$ an arbitrary set of index.

consider $f:\cup A_x\rightarrow\mathbb{R^n}$ with $f(x)=x$ so by the assumptions:disjoint and covering,i get that $f$ defines a bijection and by definition $f$ is also continuous with inverse continuous thus $f$ is a homeomorphism but as $\mathbb{R}^{n}$ is path-connected so is $\cup A_x$,but because its a union of disjoint sets i get a contradiction.

Could you enlighten me on where i did a fatal mistake on this proof? Thanks

Answer 1

When you write $\cup A_x$, you are just writing $\mathbb{R}$ in a weird way. The collection $\{A_x\}$ is not a topology on $\mathbb{R}$; your assumption that $\cup A_x$ is not path connected is fallacious.

Answer 2

It seems your contradiction is rooted in the following observation:

If $\{A_i\}_{i \in X} \subset X$ a connected topological space $A_i \cap A_j = \emptyset$, for $i \neq j$ then $\cup_{i \in X} A_i$ is not path connected under the subspace topology. Could you try to prove it? You can even take A and B to be open (as you have it here).

Answer 3

Here's how I would prove it:

Take $x_0\in X$, then $\mathbb R^n\setminus A_{x_0}$ is not open (because $\mathbb R^n$ is connected). But by hypothesis $\mathbb R^n\setminus A_{x_0}=\bigcup\limits_{x\in X\setminus\{x_0\}}A_x$, which is open since it is a union of open sets. Contradiction.

Answer 4

The fact that $\bigcup_x A_x$ is a union of disjoint sets isn’t enough to tell you that it’s not path connected: after all, $\Bbb R^n$ is the union of the disjoint sets $\{x\}$ as $x$ ranges over $\Bbb R^n$.

You can actually use plain old connectedness. Let $U$ be one of the open balls supposedly covering $\Bbb R^n$, and let $V$ be the union of the rest. Then $U$ and $V$ are disjoint, non-empty open sets whose union is the connected space $\Bbb R^n$, which is impossible.

Answer 5

Another way to prove it is that if $\emptyset \ne A\in \{A_x: x \in X\}$ and $A$ has finite radius, then $A $ and $\cup \{A_y: y\in X\land A_y\ne A\}$ are disjoint non-empty open subsets of $R^n$ whose union is $R^n,$ contradicting the connectedness of $R^n.$