#Notation Issue
18 messages · Page 1 of 1 (latest)
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I can't see the converse inclusion right now
maybe it isn't there
or maybe i made a mistake in the other inclusion as well
No I think this is right
This is most likely not true in my honest opinion
take f : {-1, 1} -> R defined by x l-> x²
and G = {{1}}
See that A = {1} will verify:
f[A] = {1}
but:
f-1[{1}] = {-1, 1} which is NOT in G
Thank you @grand vortex
Thank you @glass pike
@outer zodiac has given 1 rep to @grand vortex @glass pike
+close