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My interpretation of your question

"If

with

, and if

is not a limit point of

then does

make sense?"

Answer to that question: No. For

to exist there must be a point

such that for every

there exists a

such that

. But now suppose that

is not a limit point of

then we can pick an open neighborhood of

lets call it

such that there does not exist an element of

in that neighborhood. Now choose an value of

that requires

and we run into a contradiction.

EDIT: I am sorry, I completely for some reason disregarded the

part. This can be made slightly simpler

By the defintion of limits in

for

to exist there must exist a point

such that for every

there exsits a

such that

. But now suppose that

is not a limit point of

then there exists a neighborhood

such that there does not exist an element of

within that neighborhood. Now if we choose an

that requires

we run into a contradiction.