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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.

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  • $\begingroup$ A set is dense iff its closure is the whole space. A set is closed iff it equals its closure. So how many closed dense subsets are there in any space? $\endgroup$ – Henno Brandsma Oct 29 '15 at 14:37
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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.

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  • $\begingroup$ Right! I forgot that for e.g. $r=0.5$ the ball has only one point! I was considering only the ball I plot for the case $r=1.5$. So, if I take $r=0.5$ that is enough to prove that $A=A^o$? Regarding the second I have noticed that $d((1/n,0),(0,0)) =1$ for all $n$, but it is also true that $\lim_{n\to \infty} 1/n = 0$. So for me the sequence was convergent (but it seems that I am wrong, and I also get confused now). Finally, last but least, can you suggest any good place on the web where to find similar exercises with solution? Thanks! $\endgroup$ – Ubaldo Tiberi Oct 29 '15 at 14:56
  • $\begingroup$ Yes for the interior. The limit statement is true in the usual metric (which would correspond to most people's intuition), not in this one. Forget your intuition, apply the definitions ! Finally, I had a good course in metric topology (25 years ago or so) which had plenty of them, but those notes aren't public. Looking for specific text books on metric spaces should turn something up, I don't really know them. $\endgroup$ – Henno Brandsma Oct 29 '15 at 14:59

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