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Which one of the following spaces is not Hausdorff?

Which one of the following spaces is not Hausdorff?

Another topological space, which is not Hausdorff (unfortunately not T1, either), is the open interval (0,1) with the nested interval topology. of a infinite set X is also not Hausdorff.

Why is cofinite topology not Hausdorff?

An infinite set with the cofinite topology is not Hausdorff. In fact, any two non-empty open subsets O1,O2 in the cofinite topology on X are complements of finite subsets. Therefore, their intersection O1 \ O2 is a complement of a finite subset, but X is infinite and so O1 \ O2 6= ;. Hence, X is not Hausdorff.

Is the trivial topology Hausdorff?

In particular, it is not a Hausdorff space. Not being Hausdorff, X is not an order topology, nor is it metrizable. X is, however, regular, completely regular, normal, and completely normal; all in a rather vacuous way though, since the only closed sets are ∅ and X.

Which topologies are Hausdorff?

A Hausdorff space is a topological space with a separation property: any two distinct points can be separated by disjoint open sets—that is, whenever p and q are distinct points of a set X, there exist disjoint open sets Up and Uq such that Up contains p and Uq contains q.

Is the standard topology Hausdorff?

(a) Rn with the standard topology is a Hausdorff space. (b) R with the finite complement topology is NOT a Hausdoff space. Suppose that there are disjoint neigh- borhoods Ux and Uy of distinct two points x and y.

Is a metric topology Hausdorff?

(1.12) Any metric space is Hausdorff: if x≠y then d:=d(x,y)>0 and the open balls Bd/2(x) and Bd/2(y) are disjoint. To see this, note that if z∈Bd/2(x) then d(z,y)+d(x,z)≥d(x,y)=d (by the triangle inequality) and d/2>d(x,z), so d(z,y)>d/2 and z∉Bd/2(y).

Is finite complement topology Hausdorff?

Is the finite Complement topology on R Hausdorff? No, It is not Hausdorff.

Is every metric space Hausdorff?

(3.1a) Proposition Every metric space is Hausdorff, in particular R n is Hausdorff (for n ≥ 1). r = d(x, y) ≤ d(x, z) + d(z, y) < r/2 + r/2 i.e. r

Is indiscrete topology compact?

Exercise 3.16 – Show that any space X with the indiscrete topology is compact. Proof: Let X be a space with the indiscrete topology, that is, the only open sets in X are X and ∅. Let U = {Uα}α∈I be an open cover of X.

Is the indiscrete topology connected?

Every indiscrete space is connected. Let X be an indiscrete space, then X is the only non-empty open set, so we cannot find the disconnection of X. Hence X is connected. A subspace Y of a topological space is said to be a connected subspace if Y is connected as a topological space in its own right.

Are the rationals Hausdorff?

The Attempt at a Solution I used the following theorem: If X is a Hausdorff space, then X is locally compact iff given x in X, and a neighborhood U of x, there exists a neighborhood V of x such that Cl(V) is compact and Cl(V) is contained in U. The rationals are clearly Hausdorff.

Are all metric spaces Hausdorff?