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I want to prove that if $|J| < |I|$ where $I$ is infinite, then $|\bigcup_{j \in J}E_j| < |I|$

where $E_j \subset I$ is a finite set.

My understanding is that we need to show that there is no bijection from from $\bigcup_{j \in J}E_j$ to $I$, but there is an injection from $\bigcup_{j \in J}E_j$ to $I$.

As a starting example, one might let $I = \mathbb{R}$ and $J = \mathbb{N}$. The theorem seems true using this example, but I'm not sure where to go.

As motivation for where this might be useful, I was looking at https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Muno
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1 Answers1

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Note first that if $I$ is countable then $J$ is finite, so the LHS set is finite and you're done (since the RHS set is infinite). If $|I|>\aleph_0$, then $$\left| \bigcup_{j \in J} E_j \right| \le \sum_{j \in J} |E_j| \le \sum_{j \in J} \aleph_0 = |J| \cdot \aleph_0 = \mathrm{max} \{ |J|, \aleph_0 \} < |I|$$ Unpacking this in terms of functions is altogether more difficult, albeit not impossible; finding an injection $\bigcup_{j \in J} E_j \to J \times \mathbb{N}$ is easy enough, but then justifying that $|J| \cdot \aleph_0 < |I|$ is harder if you can't invoke some lemmas about cardinal numbers.

Clive Newstead
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  • Just as a comment for future readers: see the following if you're interested in seeing why $\max{|J|, \aleph_{0} } < |I|$. https://math.stackexchange.com/questions/690706/cardinality-of-cartesian-product-of-uncountable-set-with-countable-set – Muno Sep 26 '17 at 21:44