- #1
alex297
- 1
- 0
Homework Statement
Let X and Y be metric spaces such that X is complete. Show that if {fα(x) : α ∈ A} is a bounded subset of Y for each x ∈ X, then there exists a nonempty open subset U of X such that {fα(x) : α ∈ A, x ∈ U} is a bounded subset of Y.
Homework Equations
Definition of complete:
A metric space X is called complete if every Cauchy sequence in X converges in X.
Definition of bounded:
A set S is called bounded if ∃ x ∈ A, R>0 such that B(x,R) ⊃ S.
Baire Category Theorem:
Let {Un} be a sequence of open dense subsets of (X,d), X complete. Then
∪ Un, 1≤n<∞ is also dense.
The Attempt at a Solution
My friend and I have been working on this problem for a little while now and we're just plain stuck. I've included everything I find relevant above, including the Baire Category Theorem, which I actually don't see how it's relevant but we saw a solution that used it, even though we don't understand the solution. Here it is in case you can make sense of it (I underlined the parts I didn't follow...pretty much all of it):
Fix a point y0 ∈ Y. For each n≥1, define En to be the set of points x ∈ X such that d(fα(x), y0)≤n for all α.
Since the fα's are continuous, En is closed. By hypothesis, X is the union of the En's. By the Baire Category Theorem, some En has nonempty interior, which we take to be U.
Any help would be much appreciated! Thanks!