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## Homework Statement

Let X and Y be metric spaces such that X is complete. Show that if {f

_{α}(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 {f

_{α}(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 {U

_{n}} be a sequence of open dense subsets of (X,d), X complete. Then

∪ U

_{n}, 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 y

_{0}∈ Y. For each n≥1, define E

_{n}to be the set of points x ∈ X such that d(f

_{α}(x), y

_{0})≤n for all α.

Since the f

_{α}'s are continuous,

__E__. By hypothesis,

_{n}is closed__X is the union of the E__.

_{n}'s__By the Baire Category Theorem, some E__, which we take to be U.

_{n}has nonempty interiorAny help would be much appreciated!! Thanks!!!!