- #1
Bachelier
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##S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i ##
methinks yes because:##S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i \equiv \left\{{0,1}\right\}^\mathbb{N}##
methinks yes because:##S = \bigcup _{i=1}^{∞}\left\{{0,1}\right\}^i \equiv \left\{{0,1}\right\}^\mathbb{N}##
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