- #1

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## Homework Statement

I like to deal with equivalence relations, so I decided to skip to this exercise.

Let (X, d) be a metric space. If xn and yn are Cauchy sequences in X, define a relation xn ~ yn if d(xn, yn) --> 0. Denote the equivalence class of x with [x], and the set of all equivalence classes with Y. Define a metric D on Y with D([x], [y]) = lim d(xn, yn), n --> ∞.

## The Attempt at a Solution

(a) Show that ~ is an equivalence relation, and show that D is a well-defined metric.

Clearly, for any Cauchy sequence, xn ~ xn, xn ~ yn <==> yn ~ xn and (xn ~ yn & yn ~ zn) ==> xn ~ zn holds (the third arrived from the triangle inequality). It is also easy to check D is a metric, for example the triangle inequality (all limits are as n --> ∞): d(xn, zn) <= d(xn, yn) + d(yn, zn) ==> lim d(xn, zn) <= lim d(xn, yn) + lim d(yn, zn). The other "metric properties" are trivial. The metric D is well-defined, since limits are unique in R.

(b) Define h : X --> Y with h(x) = [(x, x, x, ...)] (i.e. h(x) is the equivalence class of the constant sequence (x, x, x, ...)). Show that h is an isometric imbedding.

Obviously for any x, y in X, d(x, y) = D(h(x), h(y)) = D([(x, x, ...)], [(y, y, ...)]) = d(x, y), so h is an isometry. Let's check for injectivity. Indeed, if x and y are distinct elements of X, y cannot possibly lie in h(x) (the equivalence class of the constant sequence) and vice versa. And further on, h : X --> h(X) is clearly a surjection.

Now, let's check continuity of h.

Let B([x], ε) = {[y] in Y : D([x], [y]) = lim d(xn, yn) < ε (as n --> ∞)} be an open ball around [x] in Y. The inverse image of this open ball under h is defined for constant sequences in Y only. And one can easily see that h^1(B) = {x in X: d(x, xn) < ε, for every n (n indexes the sequence xn)}. Clearly, this is a union of open balls around the elements xn of the sequence xn (again, notation is not quite consistent and clear, but I'll assume it's obvious what I mean), and hence open.

Now, the only thing remaining is to check that the inverse of h is continuous. Take an open ball B(x, ε) in X. We need to check that its image under h (which is the inverse of the inverse of h) is open. Since B(x, ε) = {y in X : d(x, y) < ε)}, and since h is an isomerty, we have d(x, y) = D(h(x), h(y)) < ε, so clearly h(B) = {[y] in Y : D([x], [y]) < ε}, which is an open ball in Y.

I hope these results are correct.