a) Let X be a topological space and assume that for each x ? z in X there is a continuous function f: X ? (R, ?) such that f(x) = 1 and f(z) = 0. Show that X is Hausdorff space.
Added by Nicholas R.
Close
Step 1
Since f is continuous, the preimages of open sets in (Ry U) are open in X. Let U = f^-1((0,1]) and V = f^-1([1,2)). Note that U and V are disjoint since f(x) = 1 and f(y) = 2. Now we need to show that U contains x and V contains y. Since f(x) = 1, x is in U. Show more…
Show all steps
Your feedback will help us improve your experience
Manisha Sarker and 91 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
a) Let X be a topological space and assume that for each x ≠ z in X there is a continuous function f: X → (R, μ) such that f(x) = 1 and f(z)= 0. Show that X is Hausdorff space.
Madhur L.
Let X, Y, Z be topological spaces, and let f: X → Y and g: Y → Z be continuous. Show that the function g∘f given by g∘f(x) = g(f(x)) is a continuous function from X into Z.
5_ Let X be a normal topological space, let E be a closed subset of X, and let f: E -> R be continuous. Show that f can be extended to continuous function from X to R. Hint: Let g: R -> (-1,1) be a homeomorphism, extend g ∘ f to a map from X to [-1,1], and decide what to do on the set where the extension assumes the values ±1.
Sri K.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD