1. ## Metric space subset

This is a homework problem in my algebra analysis class. I'm not really sure where to begin. I need to prove the statement below.

Y X where X is a metric space with function d. Then (Y,d) is a metric space with the same function d. How can I show that X and Y have the same metric function?

2. ## Re: Metric space subset

Originally Posted by selzer9
Y X where X is a metric space with function d. Then (Y,d) is a metric space with the same function d. How can I show that X and Y have the same metric function?
You can't show this. To show this, you to need consider the metric function of X and the metric function of Y. You can only talk about a metric function of a metric space. But, if I understand the problem correctly, Y is not a metric space by definition; it's just a subset of X. You need to show that this subset together with the metric function of X forms a metric space. And to do this, check the axioms of a metric space.

3. ## Re: Metric space subset

Originally Posted by selzer9
$\displaystyle Y\subset X$ where X is a metric space with function d. Then (Y,d) is a metric space with the same function d. How can I show that X and Y have the same metric function?
Assuming that $\displaystyle (X,d)$ is a metric space and $\displaystyle Y\subset X$ then $\displaystyle (Y,d)$ is a metric space. If you study this definition of a metric, then you see that the function $\displaystyle d$ does not loose any of its properties when applied to points of $\displaystyle Y$.

4. ## Re: Metric space subset

strictly speaking you should speak of a separate function d*, the restriction of d to YxY. this is because (in some views) the domain and co-domain of a function are part of the function's definition. in other words, it's not just "the rule" that defines a function, it's also "what you apply the rule TO", and "where the values of the function LIVE".

but...it is common practice to make no distinction between the surjective function

f:A→f(A)

and the "general function" it came from:

f:A→B. be aware of this...it can trip you up sometimes.

of course d* "is really" just d, because:

d*(y1,y2) = d(y1,y2), where on the right, we are considering the y's as elements of X.

in general, any statement of the form:

FOR ALL x,y,z in X: property (insert something here) is true, remains true even if we pick x,y and z from a SUBSET of X.

******

informally speaking, if two points lie within "an epsilon ball" in X, and both points are in Y, then they are in "an epsilon ball" in Y (but: "epsilon balls" in Y aren't always "round", parts of them might be "chopped off" because they lie in the part of X outside Y).