Topology Midterm 2 True/False

Unions of connected subsets are connected

False, draw two disjoint circles in R2.

Subspaces of Hausdorff spaces are Hausdorff

True

Every topological space is metrizable

False, T cofinite is not hausdorff, any nonhausdorff topology is not metrizable, T cobounded also works

If f:X to Y is continuous, and X is Hausdorff, then f(X) is Hausdorff

False, identity map from (X, T discrete) to (X, T indiscrete)

Intersections of connected subsets are connected

False, look at rational comb example, intersection is disjoint blobs.

Boundary(AxB) = (Boundary of A) x (Boundary of B)

False, take A=(0,1), B=(0,1), then boundary of AxB is the square, but boundary of A x boundary of B is the 4 corner points.

For A,B subsets of X, if AuB and AintersectB are connected, then both A and B are connected

False, take A=Q, B=R/Q. then R=AuB and AintersectB is empty, and R and empty are connected, but A and B are not connected.

Subspaces of connected spaces are connected

False, take Q as a subset of R. or just draw two disjoint circles in R2.

If {x} in X is closed for all x in X, then X is Hausdorff

False, take T cofinite

If f:X to Y is continuous, and T as a subset of Y is connected, then the inverse image of T is connected

False, take X to be two balls in R2, and f a constant function sending these to a single point. the single point is connected, but the inverse image is disconnected.

For S contained in A contained in X, the interior of S in A = the interior of S in X intersected with A

False, take X=R, A=Q, and S=Q. then interior of S in A is Q, interior of S in X is the empty set, empty set intersect with A is empty, not Q. or: take X=R, A=[0,1], S=[0,1/2). then S is open in A, so interior of S in A is S, but interior of S in R is (0,1/2), not S.

A function f:R to R is continuous iff its graph Gf is connected

False, take the squigly line graph example, whose function f:[0,inf) to R is defined by f(x)={0 if x=0, sin(1/x) if x is not 0} then f is not continuous but Gf is connected.

If A as a subset of X is connected, then Interior of A is also connected

False, take two balls in R2, one open one closed and intersecting at 1 point. then interior of this is two disjoint open balls, so disconnected.

If X is Hausdorff, then {x} in X is closed for all x in X

True, the complement is open