Let (X,d) be a discrete metric space i.e

d(x,y)=0 ,if x=y and d(x,y)=1 if $\displaystyle x\neq y$.

Let (Y,ρ) be any metric space

Prove that any function ,f from (X,d) to (Y,ρ) is continuous over X

Printable View

- Apr 6th 2009, 11:50 AMxalkcontinuous functions over discrete space
Let (X,d) be a discrete metric space i.e

d(x,y)=0 ,if x=y and d(x,y)=1 if $\displaystyle x\neq y$.

Let (Y,ρ) be any metric space

Prove that any function ,f from (X,d) to (Y,ρ) is continuous over X - Apr 6th 2009, 12:12 PMTheEmptySet
let $\displaystyle x_n $ be any sequence converging to x in X i.e.

$\displaystyle x_n \to x$ Using the sequential char of continuity

We need to show that $\displaystyle f(x_n) \to f(x)$ in Y

But since we are in a discreate space $\displaystyle x_n \to x$

only if there exists an N such that for all n > N $\displaystyle x_n$ is constant(why?) and $\displaystyle x_n=x$

But then for n>N $\displaystyle \rho(f(x_n),f(x))=0$ - Apr 6th 2009, 12:41 PMmylestone
Here's another approach:

$\displaystyle f: (X,d) \to (Y, \rho) $ is continuous at $\displaystyle x \in X $ if $\displaystyle \forall \varepsilon > 0, \exists \delta >0 $ such that $\displaystyle d(x,y)< \delta \Rightarrow \rho (f(x), f(y)) < \varepsilon $.

For any $\displaystyle \varepsilon >0 $, we need to let $\displaystyle \delta $ be something such that the definition above is satisfied. Let's consider the two cases:

(1) $\displaystyle 0< \varepsilon < 1 $, or

(2) $\displaystyle 1 \leq \varepsilon $.

In the first case, let's let $\displaystyle \delta = \varepsilon $. Then $\displaystyle 0< \delta < 1 $. Since we're in a discrete metric space, whenever $\displaystyle d(x,y) < \delta $, then $\displaystyle d(x,y) $ has to be 0 (because it's either 0 or 1, and by our choice of $\displaystyle \delta $ we've ruled-out the possibility of it being 1). BUT this means that $\displaystyle x=y $ (that is, the only time the distance between two points is zero is when the two points are, in fact, the same point). By definition of $\displaystyle f $ being a function, it must also be true that $\displaystyle f(x)=f(y) $. This means, of course, that $\displaystyle \rho (f(x), f(y)) =0 $. Since $\displaystyle \varepsilon $ is strictly greater than 0, it is clearly true that $\displaystyle \rho (f(x),f(y)) < \varepsilon $. Woohoo! The definition is satisfied!

The second case is easier, and I'll leave the details of that one up to you if you choose to go this route. You're going to let $\displaystyle \delta = \varepsilon $ again, and using the defining property of a discrete metric space you'll see that the definition of continuity is satisfied pretty much automatically. - Apr 6th 2009, 03:13 PMxalk
THANK YOU .

BUT considering the 2nd case,where $\displaystyle \epsilon\geq 1$ ,whether we put $\displaystyle \delta =\epsilon,\delta<\epsilon,\delta>\epsilon$,

if we have d(x,y)<δ ,then the two cases :d(x,y)=0 ,d(x,y)=1 are both satisfied.

And for the case d(x,y)=0 we can have ρ(f(x),f(y))<ε,

but for the case d(x,y) =1**,how can we have ρ(f(x),f(y))<ε??** - Apr 6th 2009, 04:38 PMmylestone
Sorry, I didn't quite think that through. I guess you can just consider $\displaystyle \varepsilon = 1 $ and $\displaystyle \varepsilon > 1 $ separately. If $\displaystyle \varepsilon =1 $, delta can be anything strictly between 0 and 1 and other than that, the same argument provided for (1) in my first reply should still work. If $\displaystyle \varepsilon > 1 $, then $\displaystyle \rho (f(x),f(y)) \leq 1 < \varepsilon, \forall f(x), f(y) $ so $\displaystyle \delta = \varepsilon $ should be fine.

That should do it, but I'm on way too much coffee and rushing out the door so if you find that I've botched it again, please let me know--thanks! - Apr 7th 2009, 02:34 AMxalk
.

The problem is with ε>1.

Ιn this case even if we put δ=ε, we cannot prove that:

d(x,y)<δ implies that ρ(f(x),f(y))<ε.Because in this case d(x,y) can be 0 and 1.

If it is zero o.k then x=y and ρ(f(x).f(y)) =0<ε,but if it is 1,how can we prove that:

ρ(f(x),f(y))< ε ?? - Apr 7th 2009, 11:12 AMmylestone
But in N with the discrete metric, $\displaystyle \rho (f(x),f(y)) $ must be less than or equal to 1 (being either 1 or 0). So if $\displaystyle \varepsilon >1 $, then it is always the case that $\displaystyle \rho (f(x),f(y)) \leq 1 < \varepsilon $. Epsilon is strictly greater than 1, and the distance between any two points of Y (in particular, any two points of the range of f) is at most--at most!--1. Does that clarify? Remember that Y has the discrete metric as well as X. $\displaystyle \delta $ can be 57,628,629,395,692,569,723,596,852 in this case if you want--the critical pieces of information here are that the distance between points in Y will never be more than one, and that epsilon is strictly greater than 1.

Hope that helps, but if not maybe someone else can provide a better explanation (sometimes I'm not the clearest-headed feller). - Apr 7th 2009, 02:03 PMxalk
- Apr 7th 2009, 07:52 PMskamoni
In a discrete metric space, any subset is clopen. Thus the inverse image

$\displaystyle f^{-1}(A)$ of any set A (subset of Y) is clopen in X. So, f is continuous. - Apr 7th 2009, 07:52 PMmylestone
HAHAAAAA--wow, that's hilarious! Funny thing is I looked back on the original post...so I misread it twice!! Man, I do this way too often. I swear I'm literate.

Let's try this again.

So we're good when $\displaystyle \varepsilon \leq 1 $ and we want to ensure that when $\displaystyle \varepsilon >1 $ we can still find a $\displaystyle \delta $ that will satisfy continuity of f. Let $\displaystyle 0< \delta \leq 1 $. Then the same comment about $\displaystyle d(x,y) < \delta \Rightarrow x=y \Rightarrow f(x) = f(y) \Rightarrow \rho (f(x),f(y)) =0 < \varepsilon $ is relevant and continuity is satisfied. You don't have to ever worry about what happens when $\displaystyle \rho (f(x),f(y)) \geq 1 $ because it'll never happen as long as you pick delta appropriately.

Alternatively, we can junk epsilons and deltas entirely if you've shown that f is continuous if and only if the preimage of an open set in Y is an open set in X. By this method, you can take an arbitrary open set in Y and no matter what it's preimage under any function is, the preimage must be open in X because every subset of a discrete metric space is open. So any function f from a discrete metric space into any other metric space is continuous.

Of course, having carried a misreading this far I wish I would have advised that route to begin with. Apologies for my misunderstanding--thanks for playing! - Apr 8th 2009, 05:11 AMxalk
So for any ε>0,we choose δ:

$\displaystyle 0<\delta\leq 1$