Metric Spaces.

[Imum Coeli]

Hello. We have just started metric spaces and I am not really sure what I'm doing. I was wondering if anyone could point out if I have made any terrible mistakes. For brevity, I'm saying that they are all complete metric spaces. Thanks.


Question: State which of the following is a metric space, giving a brief reason for your answer. For any of the them are a metric space state whether or not it is complete, giving a brief reason for your answer.

a) $\displaystyle (\mathbb{R}^n,d),$ where $\displaystyle d(x,y) := \max_i\{|x_i-y_i|\}$ where $\displaystyle x=(x_1,...,x_n), y =(y_1,...,y_n)$

b)$\displaystyle (\mathcal{P},d)$ where $\displaystyle \mathcal{P} := \{ \textit{polynomials in x}\},\; d(p_1,p_2) := \int_0^1|p_1(x)-p_2(x)|dx.$

c) $\displaystyle (\mathbb{R}^2,d)$ where $\displaystyle d((a,b),(c,d)):= |ad-bc|+|a-c|$.


Definitions:
[Metric Space]
Let X be any non-empty set. A metric on X is a function $\displaystyle d:X \times X \mapsto \mathbb{R}$ satisfying:
1) $\displaystyle d(x,y)=d(y,x) \;\forall\; x,y \in X$
2) $\displaystyle d(x,y) \geq 0 \iff x=y$
3) $\displaystyle d(x,y) \leq d(x,z)+d(z,y) \;\forall\; x,y,z \in \mathbb{R}$
The pair $\displaystyle (X,d)$ is called a metric space.

[Complete Metric Space]
A metric space is complete if every Cauchy sequence in X converges to an element of X.


Answer: (I think).

a)
1) $\displaystyle d(x,y) = \max_i\{|x_i-y_i|\} = \max_i\{|y_i-x_i|\}= d(y,x)$

2) $\displaystyle d(x,y) = \max_i\{|x_i-y_i|\}>0$ and $\displaystyle d(x,x) = \max_i\{|x_i-x_i|\}=0$

3) Since $\displaystyle x_i -y_i = x_i -z_i +z_i -y_i \implies |x_i -y_i| \leq |x_i-z_i| +|z_i -y_i|$ it follows that $\displaystyle \max_i\{|x_i-y_i|\} \leq \max_i\{|x_i-z_i|\} +\max_i\{|z_i-y_i|\}$

Hence $\displaystyle (\mathbb{R}^n,d)$ is a metric space.

Since $\displaystyle X = \mathbb{R}^n$ it is impossible to have a Cauchy sequence that converges to some $\displaystyle x_0 \notin \mathbb{R}^n$, therefore it is a complete metric space.

b)
1) $\displaystyle d(p_1,p_2) = \int_0^1|p_1(x)-p_2(x)|dx =\int_0^1|p_2(x)-p_1(x)|dx =d(p_2,p_1)$

2) Clearly $\displaystyle d(p_1,p_2) = \int_0^1|p_1(x)-p_2(x)|dx >0$ and $\displaystyle d(p_1,p_1)=\int_0^1|p_1(x)-p_1(x)|dx=0$

3) Since $\displaystyle p_1 - p_2 = p_1 -p_3 +p_3 -p_2 \implies |p_1 - p_2| \leq |p_1 -p_3| +|p_3 -p_2|$ it follows that $\displaystyle \int_0^1|p_1(x)-p_2(x)|dx \leq \int_0^1|p_1(x)-p_3(x)|dx+\int_0^1|p_3(x)-p_2(x)|dx$

Hence $\displaystyle (\mathcal{P},d)$ is a metric space.
Since the boundary of X is closed, it is complete.

c)
1) $\displaystyle d((a,b),(c,d))= |ad-bc|+|a-c| = |cb-da|+|c-a| = d((c,d),(a,b)).$

2) Clearly $\displaystyle d((a,b),(c,d))=|ad-bc|+|a-c| > 0$. Also $\displaystyle d((a,b),(a,b))= |ab-ba|+|a-a|=0$.

3) Let $\displaystyle x = (a,b), y=(c,d), z=(a,f).$ Then we need to show that $\displaystyle d((a,b),(c,d)) \leq d((a,b),(a,f))+d((a,f),(c,d)) \;\forall\; x,y,z \in \mathbb{R}$
By doing some algebra we see that this amounts to showing that $\displaystyle |ad-bc|\leq|af-ab|+|ad-cf|$.
After a bit more algebra we find (assuming I didn't make a mistake) that
$\displaystyle |ad-bc|+|bc-cf+af-ab| \leq |af-ab|+|ad-cf| \implies |ad-bc| \leq |af-ab|+|ad-cf|$

Hence $\displaystyle (\mathbb{R}^2,d)$ is a metric space.
This is also complete for similar reasons to a).

 

[HallsofIvy]

Hello. We have just started metric spaces and I am not really sure what I'm doing. I was wondering if anyone could point out if I have made any terrible mistakes. For brevity, I'm saying that they are all complete metric spaces. Thanks.


Question: State which of the following is a metric space, giving a brief reason for your answer. For any of the them are a metric space state whether or not it is complete, giving a brief reason for your answer.

a) $\displaystyle (\mathbb{R}^n,d),$ where $\displaystyle d(x,y) := \max_i\{|x_i-y_i|\}$ where $\displaystyle x=(x_1,...,x_n), y =(y_1,...,y_n)$

b)$\displaystyle (\mathcal{P},d)$ where $\displaystyle \mathcal{P} := \{ \textit{polynomials in x}\},\; d(p_1,p_2) := \int_0^1|p_1(x)-p_2(x)|dx.$

c) $\displaystyle (\mathbb{R}^2,d)$ where $\displaystyle d((a,b),(c,d)):= |ad-bc|+|a-c|$.


Definitions:
[Metric Space]
Let X be any non-empty set. A metric on X is a function $\displaystyle d:X \times X \mapsto \mathbb{R}$ satisfying:
1) $\displaystyle d(x,y)=d(y,x) \;\forall\; x,y \in X$
2) $\displaystyle d(x,y) \geq 0 \iff x=y$
3) $\displaystyle d(x,y) \leq d(x,z)+d(z,y) \;\forall\; x,y,z \in \mathbb{R}$
The pair $\displaystyle (X,d)$ is called a metric space.

[Complete Metric Space]
A metric space is complete if every Cauchy sequence in X converges to an element of X.


Answer: (I think).

a)
1) $\displaystyle d(x,y) = \max_i\{|x_i-y_i|\} = \max_i\{|y_i-x_i|\}= d(y,x)$

2) $\displaystyle d(x,y) = \max_i\{|x_i-y_i|\}>0$ and $\displaystyle d(x,x) = \max_i\{|x_i-x_i|\}=0$

3) Since $\displaystyle x_i -y_i = x_i -z_i +z_i -y_i \implies |x_i -y_i| \leq |x_i-z_i| +|z_i -y_i|$ it follows that $\displaystyle \max_i\{|x_i-y_i|\} \leq \max_i\{|x_i-z_i|\} +\max_i\{|z_i-y_i|\}$

Hence $\displaystyle (\mathbb{R}^n,d)$ is a metric space.

Looks good so far.

Since $\displaystyle X = \mathbb{R}^n$ it is impossible to have a Cauchy sequence that converges to some $\displaystyle x_0 \notin \mathbb{R}^n$, therefore it is a complete metric space.

This is not sufficient- it assumes that a Cauchy sequence, as defined with this metric does converge.

b)
1) $\displaystyle d(p_1,p_2) = \int_0^1|p_1(x)-p_2(x)|dx =\int_0^1|p_2(x)-p_1(x)|dx =d(p_2,p_1)$

2) Clearly $\displaystyle d(p_1,p_2) = \int_0^1|p_1(x)-p_2(x)|dx >0$ and $\displaystyle d(p_1,p_1)=\int_0^1|p_1(x)-p_1(x)|dx=0$

3) Since $\displaystyle p_1 - p_2 = p_1 -p_3 +p_3 -p_2 \implies |p_1 - p_2| \leq |p_1 -p_3| +|p_3 -p_2|$ it follows that $\displaystyle \int_0^1|p_1(x)-p_2(x)|dx \leq \int_0^1|p_1(x)-p_3(x)|dx+\int_0^1|p_3(x)-p_2(x)|dx$

Hence $\displaystyle (\mathcal{P},d)$ is a metric space.
Since the boundary of X is closed, it is complete.

c)
1) $\displaystyle d((a,b),(c,d))= |ad-bc|+|a-c| = |cb-da|+|c-a| = d((c,d),(a,b)).$

2) Clearly $\displaystyle d((a,b),(c,d))=|ad-bc|+|a-c| > 0$. Also $\displaystyle d((a,b),(a,b))= |ab-ba|+|a-a|=0$.

3) Let $\displaystyle x = (a,b), y=(c,d), z=(a,f).$ Then we need to show that $\displaystyle d((a,b),(c,d)) \leq d((a,b),(a,f))+d((a,f),(c,d)) \;\forall\; x,y,z \in \mathbb{R}$
By doing some algebra we see that this amounts to showing that $\displaystyle |ad-bc|\leq|af-ab|+|ad-cf|$.
After a bit more algebra we find (assuming I didn't make a mistake) that
$\displaystyle |ad-bc|+|bc-cf+af-ab| \leq |af-ab|+|ad-cf| \implies |ad-bc| \leq |af-ab|+|ad-cf|$

Hence $\displaystyle (\mathbb{R}^2,d)$ is a metric space.
This is also complete for similar reasons to a).

 

[Imum Coeli]

Okay cool. Thanks.

So for:

a)
Metric Space - Yes.
Complete - Yes

Reason:
Suppose that $\displaystyle n=2$
Then $\displaystyle |x_i -x_j|$ and $\displaystyle |y_i-y_j|,\; i,j=1,...,n \in \mathbb{N}$ are both Cauchy and both converge to some element in $\displaystyle \mathbb{R}^2$. This result applies for any $\displaystyle n\in \mathbb{N}$.
Hence $\displaystyle (\mathbb{R}^n,d)$ is complete.

b)
Metric Space - Yes.
Complete - No.

Take $\displaystyle p1(x) =$ nth order Taylor polynomial for exp(x) and $\displaystyle p_2(x)=$ the zero polynomial. Then $\displaystyle p1-p2$ is Cauchy and converges to exp(x) which is not a polynomial. Hence $\displaystyle \exists$ a Cauchy sequence that does not converge to a point in X.
Therefore $\displaystyle (\mathcal{P},d)$ is not complete.

c)
Metric Space - No.
Complete - N/A

In condition 2) of the definition for a metric space, suppose that $\displaystyle a=c=0$ and $\displaystyle b \neq d \in \mathbb{R}$. Then $\displaystyle d(0,b),(0,d))=|0d-b0|+|0-0| = 0$ but $\displaystyle (0,b)\neq (0,d)$
So $\displaystyle (\mathbb{R}^2,d)$ is not a metric space.