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.