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A similar question was asked here (Is every open subset of $ \mathbb{R} $ uncountable?) and here (Question on the existence of finite open subsets in $\mathbb{R}^{k}$).

I know nothing of rigorous cardinality, so a proof without that concept would be appreciated (perhaps a modified version of the one found in the second link). If the proof requires cardinality, could you explain why the cardinality of an open subset is the same as $\mathbb{R}^n$: perhaps there exists a bijection we can show?

J.-E. Pin
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Ben
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  • I'm confused... how is your question distinct from the two to which you have linked? What new question are you asking? If you are asking about uncountability, you are automatically asking about something which involves cardinality... – Xander Henderson Sep 17 '19 at 00:27

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Let $x$ be a point in an open set $U$ in $\mathbb R^{n}$. Then there exists $r>0$ such that $x+t(1,0,0,...,0) \in U$ for $|t| < r$. Define a map $f:(-r,r) \to U$ by $f(t)=x+t(1,0,0,...,0)$. Verify that this maps is one-to-one. What can you conclude from this?

Kavi Rama Murthy
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