Let "(X, C)" be a topological space, where "C" is cofinite topology on "X". Let us show that "N(p)\\subseteq C" where "N(p)" is set of all neignborhoods of "p\\in X".
By defenition, a subset "U\\subseteq X" is called an open set of topological space "(X, C)" if "U" is an element of topology "C". A neignborhood of a point "p" is an open set "U" such that "p\\in U". It follows that each neignborhood of "p" belongs to "C". Therefore, "N(p)\\subseteq C".
On the other hand, the set "V=X\\setminus\\{p\\}" is an open set in cofinite topology "C" on "X", and "p\\notin V". Therefore, "N(p)\\ne C."
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