Exact(24)
(X, d) is properly amenable.
Suppose now that X is amenable but not properly amenable.
Hence, X is amenable but not properly amenable.
One can easily show that, if X is properly amenable, then (Y_2) is also properly amenable, contradicting our hypothesis.
If (Y_2) is amenable, then it is also properly amenable by Proposition 2.18, and so X is also properly amenable, contradicting our hypothesis.
Then X is amenable if and only if X is properly amenable.
Similar(36)
We claim that ({mathcal A}) is left properly algebraically amenable but not right algebraically amenable.
Let E be the following graph: Open image in new window Here we also have that the path algebra (mathcal A : = mathbb {K}E) is left properly algebraically amenable but not right algebraically amenable, despite the existence of an exclusive maximal cycle.
X is amenable if X is left and right amenable.
(2) (widetilde{{mathcal A}}) is properly algebraically amenable if and only if ({mathcal A}) is properly algebraically amenable. .
(widetilde{{mathcal A}}) is properly algebraically amenable if and only if ({mathcal A}) is properly algebraically amenable.
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