Metric spaces exercise: draw a ball, check the convergence of a sequence, find the interior of a set and the dense subsets.

Question

To assess my basic understanding of metric spaces, I am doing the following exercise. However, I don't know if it is correct.

Let $X=\mathbb{R}^2$ and let $d(x,y)$ the Hamming distance, i.e. $d(x,y)$ is number of places where $x$ and $y$ have different entries.

a) Sketch $\mathcal{B}_{0.5}((0,0))$ and $\mathcal{B}_{1.5}((0,0))$;

b) Does the sequence $x_n=(1/n,0)$ converges to zero?

c) let $A:=\{(x,y): 0\le x \le 1, 0\le y \le 1\}$. Find $A^o$;

d) Find all the dense set in $X$.

My Solution

a) Given that we are in $\mathbb{R}^2$, then $d(x,y)=\{0,1,2\}$, i.e. two vector can have at most two different entries. In order to design the ball, I have to find the set of points for which $d((0,0),(x,y))$ is less than 0.5 and less than 1.5. Here are the two balls that I have found Ball of radius 0.5 Ball of radius 1.5

b) $lim_{n\to\infty} d((1/n,0),(0,0)) = 0$. The sequence converges.

c) $A^o = \{(x,y): 0< x < 1, 0< y < 1\}$. Infact, any ball centered in $(0,0)$ contains some points with coordinate $x<0$ and $y<0$ (thus they belong to the ball but not to the set $A$). A ball in $(0,1)$ contains some point $y>1$ and a ball in $(1,0)$ contains some point $x>1$.

d) Every closed subset of $\mathbb{R}^2$ is dense.

Answer 1

Your sketches of the balls seem correct. In formulae: $\mathcal{B}_{0.5}((0,0)) = \{(0,0\}$ and $\mathcal{B}_{1.5}((0,0)) = \{(x,y): x = 0 \text{ or } y = 0\}$.

The second: $d((\frac{1}{n},0),(0,0)) = 1$ for all $n$. So?

As to the third: for any point, the ball with radius $\frac{1}{2}$ will only contain that point. This makes your reasoning false. In fact $A^\circ$ is defines as all $\{x \in X: \exists r> 0: B_r(x) \subseteq A \}$. Take $r = \frac{1}{2}$ always. What do you conclude?

A set $D$ is dense iff for every $x$ and for every $r > 0$, $B_r(x)$ intersects $D$. Consider $r = \frac{1}{2}$ again.