- #1

WWGD

Science Advisor

Gold Member

- 5,998

- 7,192

First, we see that intervals [c,d] ; ##0\geq c< d\geq 1## are open, by intersecting intervals ##[a,x) \cap [0,1]##. It then follows that ##[0,1]_New=[0,a) \cup [a,1]## is disconnected. Hausdorff is relatively straightforward. Given## a,b \in [0,1] ; 0<a<b<1## Then we can use ##[0,a], [b,1]##( some accomodations when b=1)

It is compactness that seems harder. We have that X_new is strictly finer than X, since we saw (a,b) is open in X_new. Then the identity map from X_new to X is continuous, a continuous bijection.

Then all I have to show [0,1] is not compact in X_new is an argument by contradiction: A continuous bijection between compact and Hausdorff is a homeomorphism, which cannot happen, because the topologies X, X_ new are different . But this seems too high-powered an argument. Is there a simpler way of proving [0,1]_ new is not compact?