# uniform continuity of x

#### cp05

is f(x)=x from (r, normal) to (r, discrete) uniformly continuous?

my first guess is yes...because letting some delta=epsilon, d(u,v)<delta
and d(f(u),f(v))=d(u,v)<delta=epsilon
so it seems uniformly continuous...

#### Drexel28

MHF Hall of Honor
is f(x)=x from (r, normal) to (r, discrete) uniformly continuous?

my first guess is yes...because letting some delta=epsilon, d(u,v)<delta
and d(f(u),f(v))=d(u,v)<delta=epsilon
so it seems uniformly continuous...
Do you mean with the discrete metric? I don't even see why it's continuous. IF it were continuous then $$\displaystyle f^{-1}(\{0\})=\{0\}$$ must be open in $$\displaystyle \mathbb{R}$$ with the usual metric.

#### Drexel28

MHF Hall of Honor

#### cp05

sorry but I don't understand any of the explanations

isn't the image of x in the discrete metric just the constant function 1? because all of the x-values are different from each other?

#### mabruka

to show continuity you must show that the inverse image of an open set in (r,discrete) is open in (r,normal) right?

So take for example the set {a}. This set is open in (r,discrete). while the inverse image is

$$\displaystyle f^{-1}(a)=\{x\in \mathbb R: f(x)=a\}=\{a\}$$ which clearly is not open in (r,normal).

so f is cant be continuous wi those two topologies.

Take note that changing the topologies doesnt change the way the function acts, or the inverse image acts, the crucial point is whether the set and its inverse image are open or not.

#### cp05

how do you know {a} is open? isn't it a singleton so it has to be closed?

#### mabruka

Thats exactly what i said, that {a} is not open!! read it again! We know what singletons are not open sets, since they fail to satisfy the open set definition.

In fact, as you point it out, it is closed (although this alone doesnt mean it is not open, for example all $$\displaystyle \mathbb R$$ is both open and closed.)

#### cp05

I meant, why is {a} open in (R, discrete)?

#### Drexel28

MHF Hall of Honor
I meant, why is {a} open in (R, discrete)?
That's pretty much the definition. Note that $$\displaystyle B_{\frac{1}{2}}(a)\subseteq\{a\}$$ and thus $$\displaystyle a$$ is an interior point of $$\displaystyle \{a\}$$ and since that is the only point of $$\displaystyle \{a\}$$ it follows that every point of $$\displaystyle \{a\}$$ is an interior point...soo