uniform continuity of x

Apr 2010
43
1
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
Nov 2009
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Berkeley, California
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.
 
Apr 2010
43
1
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?
 
Jan 2010
150
29
Mexico City
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.
 
Apr 2010
43
1
how do you know {a} is open? isn't it a singleton so it has to be closed?
 
Jan 2010
150
29
Mexico City
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.)
 
Apr 2010
43
1
I meant, why is {a} open in (R, discrete)?
 

Drexel28

MHF Hall of Honor
Nov 2009
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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