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Problem. Let $\Lambda$ be an index set. Then show that $$\displaystyle\bigcup_{\alpha\in\Lambda}\left(\displaystyle\bigcup_{B\in\gamma_{\alpha}}B\right)=\displaystyle\bigcup_{B\in\gamma}B$$where $\gamma=\displaystyle\bigcup_{\alpha\in\Lambda}\gamma_{\alpha}$.

This exercise came up when I was trying to solve a problem of topology. Although I think that the result is "intuitively obvious" to me, I can't write a purely formal proof of this claim.

Basically what we need to prove is,

$$x\in \displaystyle\bigcup_{\alpha\in\Lambda}\left(\displaystyle\bigcup_{B\in\gamma_{\alpha}}B\right)\iff(\exists\alpha_{0}\in\Lambda)\left[x\in\displaystyle\bigcup_{B\in\gamma_{\alpha_0}}B\right]\iff x\in \displaystyle\bigcup_{B\in\gamma}B$$

More precisely, so far as I can see, the first "$\iff$" follows from definition (correct me if I am wrong) but to me writing the proof of the second "$\iff$" doesn't seem so easy (in fact, as I said earlier I can't write a formal proof). Can anyone help?

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$$(\exists\alpha_{0}\in\Lambda)\left[x\in\displaystyle\bigcup_{B\in\gamma_{\alpha_0}}B\right] \iff (\exists \alpha_0 \in \Lambda)[(\exists B \in \gamma_{\alpha_0})(x \in B)] \iff [(\exists \alpha_0 \in \Lambda)(\exists B \in \gamma_{\alpha_0})](x \in B) \iff \left(\exists B \in \bigcup_{\alpha \in \Lambda} \gamma_{\alpha} \right)(x \in B) \iff (\exists B \in \gamma)(x \in B) \iff x \in \bigcup_{B \in \gamma} B $$