Family of continuous functions defined on complete metric spaces

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
alex297
Messages
1
Reaction score
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!
 
Physics news on Phys.org
alex297 said:

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!

Let's take your problems one at a time. Start with the statement [itex]E_n[/itex] is closed. For a fixed α is the set of points x ∈ X such that d(fα(x), y0)≤n closed? What's the relation of [itex]E_n[/itex] to those sets??