Let INF={<M>|L(M) is infinite}.  Show that INF is not semi-decidable.

Solution:  to show that INF is not semi-decidable we show that
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Atm is many-one reducible to INF:
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Thus, given an input <M,w> to Atm, we must construct an input <M'> to
INF such that the following condition of correctness is satisfied:

M does not accept w <==> L(M') is infinite.

Design M' as follows: on input x, M' simulates M on w for |x| many
steps, and M' accepts iff M did not accept w within these |x| steps.

Now, give a proof of correctness to understand why this reduction
works.