So, I want to prove this here $F(A) = \bigcup_{A'\subseteq A}F(A')$, where $A'$ runs through all finite subsets of $A$. $F$ is a field and $A$ is some subset of it.

My argument is as follows:

"$\supseteq$": $F(A) \supseteq F(A') \;\;\forall\; A'\subseteq A \implies F(A) \supseteq \bigcup_{A'\subseteq A}F(A')$

"$\subseteq$": Let $X\subseteq F(A)$ be a finite subset. Then there is a finite subset $A''\subseteq A$ with $X\subseteq F(A'')\subseteq \bigcup_{A'\subseteq A}F(A')$. Therefore, since $X$ is arbitrary, it follows $F(A)\subseteq \bigcup_{A'\subseteq A}F(A')$.

Is this correct?

I'm worried that I need to flesh out the part where I conclude that there is a finite subset $A''\subseteq A$ with $X\subseteq F(A'')$, but this already follows from $X$ being a finite subset of $F(A)$ and the definition of $F(A'')$, right?

  • $\begingroup$ What's $F(A)$? ${}{}{}$ $\endgroup$ – user26857 Dec 17 '16 at 15:42
  • $\begingroup$ The smallest field that contains F and A. $\endgroup$ – Miriam Dec 17 '16 at 15:52
  • $\begingroup$ For the other inclusion just note that $A\subseteq A$, so $F(A) \subseteq \bigcup_{A'\subseteq A}F(A')$ $\endgroup$ – Xam Dec 19 '16 at 17:24

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