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Sep 10, 2013 at 16:14 vote accept Vahideh Bagheri
Sep 6, 2013 at 15:37 comment added Ramiro de la Vega @VahidehBagheri: If $x$ is a weak $P$-point then $x$ is not in the closure of any countable $A \subseteq X \setminus \lbrace x \rbrace$ and therefore (using countable tightness) it is not in the closure of $X \setminus \lbrace x \rbrace$; that is, $x$ is isolated. The other implications: Isolated $\Rightarrow$ $P$-point $\Rightarrow$ weak $P$-point are obviously true in any space (never mind the tightness).
Sep 6, 2013 at 15:04 comment added Vahideh Bagheri [quote]First note that in a space with countable tightness: $P$-point $\Leftrightarrow$ weak $P$-point $\Leftrightarrow$ Isolated point.[\quote] Dear Ramiro de la Vega, I don't find above proposition in literatures. please refer me to that refrense.
Sep 2, 2013 at 8:07 comment added Vahideh Bagheri Thank u dear. This example was that I wish. Do you know QP-spaces? (spaces X that C(X) has no prime z-ideal except primes and maximals.) This space is a poinwise QP-space.
Aug 22, 2013 at 16:02 history answered Ramiro de la Vega CC BY-SA 3.0