## Urysohn’s Lemma for Metric Spaces.

### November 20, 2011

After proving this "Deep Result" [cf. Munkres] a professor will (hopefully) say something like: "Yes, the proof is important, but what does this theorem mean?  And what does it mean for spaces which are sufficiently nice, like metric spaces?"

Let’s state the result just so we’re all on the same page.

Theorem.   The topological space $X$ is normal (that is, every pair of disjoint closed subsets $A, B$ can separated by disjoint neighborhoods) if and only if any two disjoint closed sets can be separated by a function.

The proof of this is not obvious.  In fact, it is quite involved.  But if $X$ happens to be a metric space, we can make the proof significantly easier by using a function that is similar to a distance function with a few minor modifications.  I call this function the standard Urysohn function for metric spaces for lack of a better (or shorter) name.  The function is as follows:

$\displaystyle f(x) = \frac{d(x,A)}{d(x,A) + d(x,B)}$

but before proving this theorem in the metric space setting, let’s look at this function.  If $A,B$ are disjoint, then it is clear that the denominator is non-zero (why?) so this is defined everywhere.  If $x\in A$ we have that this function evaluates to 0.  If $x\in B$ then we have that this function evaluates to 1.  And the function (since the distance is always non-negative) achieves values in $(0,1)$ for every point not in $A$ or $B$ (convince yourself that this function is continuous and only achieves the values 0 and 1 if the point is in either $A$ or $B$ respectively).  This function, therefore, separates $A$ and $B$

The next thing you should think about is: what do the preimages of $f$ look like?  Draw some pictures!  Here’s some pictures to start you off to get the idea.  The first is just in ${\mathbb R}$ and the second is in the real plane.  Both have the standard topology.

This one is potentially not to scale, but you’ll see that we essentially have a linear relation (since the sum in the denominator will always be 1 in this case) except when we start looking at points that are less than all the points in $A$ or greater than all the points in $B$.  What happens in those cases?

I’ve left this "face" as an exercise because when I did it I was kind of excited about the result.  What do the preimages look like?  Does this look like something you’ve seen before; maybe in physics?  Could you modify the Urysohn equation above to make it seem MORE like something in physics?

So now that you’ve seen these pre-images, it should be relatively clear how to create neighborhoods around each of the open sets.  It still takes a bit of proof, but it’s nowhere near the difficulty of the standard Urysohn’s.

I encourage you, further, to draw some pictures.  My rule of thumb is: draw at least five different pictures; four easy ones and a hard one.  Find out where the preimages are in each of these.  Remember, too, that not every metric space looks the same; what would this look like, for example, in the taxi-cab space?  Or in some product space?  Or in the discrete space…?

### One Response to “Urysohn’s Lemma for Metric Spaces.”

1. A. said

It is important to also show why d(x,A) is continuous for every closed subset A. While it is true that d(x,y) is continuous for every y, don’t forget that d(x,A) is an infimum, which doesn’t necessarily preserve continuity.