Now suppose "f": N → "X" is a bijection, and let {"xi" : "i" ∈ N} denote the image of "f". Let "X" be the first open set and choose a neighbourhood "U"1 ⊂ "X" whose closure does not contain "x"1. Secondly, choose a neighbourhood "U"2 ⊂ "U"1 whose closure does not contain "x"2. Continue this process whereby choosing a neighbourhood "U""n"+1 ⊂ "Un" whose closure does not contain "x""n"+1. Then the collection {"Ui" : "i" ∈ N} satisfies the finite intersection property and hence the intersection of their closures is non-empty by the compactness of "X". Therefore, there is a point "x" in this intersection. No "xi" can belong to this intersection because "xi" does not belong to the closure of "Ui". This means that "x" is not equal to "xi" for all "i" and "f" is not surjective; a contradiction. Therefore, "X" is uncountable.