# Implications.

• Jan 26th 2010, 03:18 PM
Showcase_22
Implications.
Quote:

Is at least one of the following two implications valid for every map $\displaystyle f:M \rightarrow N$ of a metric space $\displaystyle M$ to a metric space $\displaystyle N$?

a). $\displaystyle f$ is continuous $\displaystyle \Rightarrow \ f(U)$ is open for every $\displaystyle U \subset M$.

b). $\displaystyle f(U)$ is open for every open $\displaystyle U \subset M \Rightarrow \ f$ is continuous.
I think that a). is valid. The definition of continuity, at a point $\displaystyle c \in M$, adapted to this case is:

$\displaystyle (\forall \varepsilon >0)(\exists \delta >0)[d_M(c,x)< \delta \ \Rightarrow \ d_N(f(c),f(x))< \varepsilon]$

$\displaystyle d_N(f(c),f(x))< \varepsilon$ is just $\displaystyle B_{\varepsilon}(f(c),d_N)$ valid $\displaystyle \forall c,x \in M$.

Hence it does imply that $\displaystyle f(U)$ is open for every $\displaystyle U \subset M$.

So, i've answered the question with a yes! However, for the sake of curiosity I will try and figure out b).

Since a). is true, i'll try and find a counterexample.

Consider the French Railway Metric. This is open because a ball can be constructed around every point.

However, $\displaystyle f$ is not continuous.
Suppose that $\displaystyle x=(1,0)$ and $\displaystyle x_n=(1+\frac{1}{n},0)$.

As $\displaystyle d(x,x_n) \rightarrow 0$, $\displaystyle d_F(x,x_n) \rightarrow 2$ so it cannot be continuous.

Is this the correct way to think about the problem?
• Jan 26th 2010, 05:45 PM
Defunkt
Quote:

Originally Posted by Showcase_22
I think that a). is valid. The definition of continuity, at a point $\displaystyle c \in M$, adapted to this case is:

$\displaystyle (\forall \varepsilon >0)(\exists \delta >0)[d_M(c,x)< \delta \ \Rightarrow \ d_N(f(c),f(x))< \varepsilon]$

$\displaystyle d_N(f(c),f(x))< \varepsilon$ is just $\displaystyle B_{\varepsilon}(f(c),d_N)$ valid $\displaystyle \forall c,x \in M$.

Hence it does imply that $\displaystyle f(U)$ is open for every $\displaystyle U \subset M$.

How? To prove that $\displaystyle f(U)$ is open, you need to take a point $\displaystyle n \in f(u)$ and show that there exists an $\displaystyle \epsilon_n > 0$ such that $\displaystyle B_{d_N}(n, \epsilon_n) \subseteq f(U)$. Have you done that?

After you understand why your proof is wrong, think of $\displaystyle M=N=(\mathbb{R}, d_2)$ with $\displaystyle f(x)=x$.

Quote:

Consider the French Railway Metric. This is open because a ball can be constructed around every point.

However, $\displaystyle f$ is not continuous.
Suppose that $\displaystyle x=(1,0)$ and $\displaystyle x_n=(1+\frac{1}{n},0)$.

As $\displaystyle d(x,x_n) \rightarrow 0$, $\displaystyle d_F(x,x_n) \rightarrow 2$ so it cannot be continuous.

Is this the correct way to think about the problem?
What is open? You haven't specified any space, just the metric!
• Jan 26th 2010, 06:12 PM
Drexel28
Quote:

a) If $\displaystyle f:\left(X,d\right)\mapsto\left(\bar{X},\bar{d}\rig ht)$ is continuous, then it's open.
Take $\displaystyle X=\bar{X}=\mathbb{R}$ and the usual metric and consider $\displaystyle f(x)=0$

Quote:

If a function is open, it is continuous.
It is relatively easy to prove that if $\displaystyle f$ is bijective and open then $\displaystyle f^{-1}$ is continuous. Thus, to state the above would imply that every bijective function with continuous inverse is continuous. Let $\displaystyle g:[0,1]\cup\left(2,3\right]\mapsto[0,2]$ be such that $\displaystyle x\mapsto\begin{cases} x & \mbox{if} \quad x\in [0,1] \\ x-1 & \mbox{if} \quad x\in (2,3]\end{cases}$. This is a bijection which is continuous but has a non-continuous inverse. So...
• Jan 26th 2010, 07:56 PM
Showcase_22
Quote:

Take and the usual metric and consider
$\displaystyle f:(\mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})$

$\displaystyle f(x)=0$ maps every point to 0. The range is just the set $\displaystyle \{ 0 \}$ which is closed.

$\displaystyle f(x)=0$ is continuous so we have a counterexample to the question!

Quote:

every bijective function with continuous inverse is continuous.
The function you've written contradicts the above quote!
• Jan 26th 2010, 08:01 PM
Drexel28
Quote:

Originally Posted by Showcase_22
$\displaystyle f:(\mathbb{R},d) \rightarrow (\mathbb{R}, \overline{d})$

$\displaystyle f(x)=0$ maps every point to 0. The range is just the set $\displaystyle \{ 0 \}$ which is closed.

$\displaystyle f(x)=0$ is continuous so we have a counterexample to the question!

The function you've written contradicts the above quote!

Yes. Both are wrong. I was providing counterexamples to my quotes.
• Jan 26th 2010, 08:15 PM
Showcase_22
Thankyou very much!(Rofl)