Implications.

• Jan 26th 2010, 04:18 PM
Showcase_22
Implications.
Quote:

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

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

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

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

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

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

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

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

Is this the correct way to think about the problem?
• Jan 26th 2010, 06:45 PM
Defunkt
Quote:

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

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

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

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

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

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

Quote:

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

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

As $d(x,x_n) \rightarrow 0$, $d_F(x,x_n) \rightarrow 2$ 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, 07:12 PM
Drexel28
Quote:

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

Quote:

If a function is open, it is continuous.
It is relatively easy to prove that if $f$ is bijective and open then $f^{-1}$ is continuous. Thus, to state the above would imply that every bijective function with continuous inverse is continuous. Let $g:[0,1]\cup\left(2,3\right]\mapsto[0,2]$ be such that $x\mapsto\begin{cases} x & \mbox{if} \quad x\in [0,1] \\ x-1 & \mbox{if} \quad x\in (2,3]\end{cases}$. This is a bijection which is continuous but has a non-continuous inverse. So...
• Jan 26th 2010, 08:56 PM
Showcase_22
Quote:

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

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

$f(x)=0$ 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, 09:01 PM
Drexel28
Quote:

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

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

$f(x)=0$ 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, 09:15 PM
Showcase_22
Thankyou very much!(Rofl)