shah_nosrat Posted March 4, 2012 Share Posted March 4, 2012 Here the question I need to prove followed by my attempt at the solution; [math] \bigcup_{\beta \in \mathcal{B}} A_{\beta} \subseteq \bigcup_{\alpha \in \mathcal{A}} A_{\alpha} [/math] and suppose [math] \mathcal{B} \subseteq \mathcal{A} [/math] My attempt at the solution, as follows: Let [math] x \in \bigcup_{\beta \in \mathcal{B}} A_{\beta} [/math] such that for some [math] (\beta \in \mathcal{B}) [/math] we have [math] x \in A_{\beta} [/math]. Now, Pick [math] \beta \in \mathcal{B} [/math] , since [math] \mathcal{B} \subseteq \mathcal{A} [/math] we have [math] \beta \in \mathcal{A} [/math]. Hence we have [math] x \in A_{\alpha} [/math] for some [math] \alpha [/math]. Which follows: [math] x \in \bigcup_{\alpha \in \mathcal{A}} A_{\alpha} [/math] Is the above proof correct? Link to comment Share on other sites More sharing options...
mathematic Posted March 4, 2012 Share Posted March 4, 2012 Looks good! Link to comment Share on other sites More sharing options...
DrRocket Posted March 4, 2012 Share Posted March 4, 2012 Is the above proof correct? yes 1 Link to comment Share on other sites More sharing options...
shah_nosrat Posted March 5, 2012 Author Share Posted March 5, 2012 Thank you! Link to comment Share on other sites More sharing options...
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