Which of the following metric spaces is complete?

1. $X_1=(0,1), d(x,y)=|\tan x-\tan y|$

2. $X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}$

3. $X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$

4. $X_4=\mathbb{R}, d(x,y)=|e^x-e^y|$

The numbers $\frac{1}{2^n}-\frac{1}{2^m}$ and $\frac{1}{2^n}$ are close to zero for large $m,n$ in the usual metric and hence the numbers $tan(\frac{1}{2^n}-\frac{1}{2^m})$ and $tan(\frac{1}{2^n})$ are also close to zero in the usual metric because $\lim _{\theta \to 0} (tan \theta) = 0$. This shows that, from the above equation, $\{\frac{1}{2^n}\}$ is a cauchy sequence in $X_1$.

We have $\{\frac{1}{2^n}\}$ is a decresing sequence and hence if it converge in $(X_1,d)$ it should converge to 0. Since $0$ is not an element of $X_1 = (0,1)$, the sequence $\{\frac{1}{2^n}\}$ is not converging in $X_1$. Therefore $X_1$ is not complete

Claim: Discrete metric spaces are complete.

Proof:

Let $\{x_n\}$ be a Cauchy sequence in a discrete metric space $(X,d)$, then given $\epsilon > 0$ there exists a positive integer $p \in \Bbb{N}$ such that $d(x_n , x_m) < \epsilon$ for all $m,n \ge p$. Let $\epsilon = 1/2$ then there exists a positive integer $p \in Bbb{N}$ such that $d(x_n , x_m) < 1/2$ for all $m,n \ge p$. But $d(x_n, x_m) < 1/2 \to d(x_n , x_m) = 0$ $m,n \ge p$ in discrete metric. Hence $x_n = x_m = a (say)$ for all $m,n \ge P$ and hence this cauchy sequence $\{x_n\}$ is an eventually constant sequence $a$ and converge to $a$. This completes the proof.

4. (

We have for $k \le n$, $$d(-k,-n)=|e^{-k}-e^{-n}|=e^{-k}-e^{-n}<e^{-k}\le e^{-m_\epsilon}<\epsilon$$

Similarly for $n \le k$, $$d(-k,-n)<\epsilon$$ by the symmetric property of the metric. This shows that the sequence $\{-n\}$ in $X_4$ is a cauchy sequence and it is clearly not convergent as $-\infty$ is not in $\Bbb{R}$.

1. $X_1=(0,1), d(x,y)=|\tan x-\tan y|$

2. $X_2=[0,1], d(x,y)=\frac{|x-y|}{1+|x-y|}$

3. $X_3=\mathbb{Q}, d(x,y)=1\forall x\neq y$

4. $X_4=\mathbb{R}, d(x,y)=|e^x-e^y|$

**Solution :****1. (**

**not complete**) Consider the sequence $\{\frac{1}{2^n}\}$ in $X_1$. We claim that this is a Cauchy sequence which does not converge in $(X_1,d)$. We have $$d(\frac{1}{2^n},\frac{1}{2^m}) = |tan(\frac{1}{2^n})-tan(\frac{1}{2^m})| = |tan(\frac{1}{2^n}-\frac{1}{2^m})(1-tan(\frac{1}{2^n})tan( \frac{1}{2^m}))|.$$The numbers $\frac{1}{2^n}-\frac{1}{2^m}$ and $\frac{1}{2^n}$ are close to zero for large $m,n$ in the usual metric and hence the numbers $tan(\frac{1}{2^n}-\frac{1}{2^m})$ and $tan(\frac{1}{2^n})$ are also close to zero in the usual metric because $\lim _{\theta \to 0} (tan \theta) = 0$. This shows that, from the above equation, $\{\frac{1}{2^n}\}$ is a cauchy sequence in $X_1$.

We have $\{\frac{1}{2^n}\}$ is a decresing sequence and hence if it converge in $(X_1,d)$ it should converge to 0. Since $0$ is not an element of $X_1 = (0,1)$, the sequence $\{\frac{1}{2^n}\}$ is not converging in $X_1$. Therefore $X_1$ is not complete

**2. (**

**complete**) because the given metric is equivalent to the usual metric on $[0,1]$.**3. (**

**complete**) because it is a discrete metric space.Claim: Discrete metric spaces are complete.

Proof:

Let $\{x_n\}$ be a Cauchy sequence in a discrete metric space $(X,d)$, then given $\epsilon > 0$ there exists a positive integer $p \in \Bbb{N}$ such that $d(x_n , x_m) < \epsilon$ for all $m,n \ge p$. Let $\epsilon = 1/2$ then there exists a positive integer $p \in Bbb{N}$ such that $d(x_n , x_m) < 1/2$ for all $m,n \ge p$. But $d(x_n, x_m) < 1/2 \to d(x_n , x_m) = 0$ $m,n \ge p$ in discrete metric. Hence $x_n = x_m = a (say)$ for all $m,n \ge P$ and hence this cauchy sequence $\{x_n\}$ is an eventually constant sequence $a$ and converge to $a$. This completes the proof.

4. (

**not complete**) Consider the sequence $\{-n\}$ in $X_4$. We claim that this is a Cauchy sequence that does not converge in $(X_4,d)$.We have for $k \le n$, $$d(-k,-n)=|e^{-k}-e^{-n}|=e^{-k}-e^{-n}<e^{-k}\le e^{-m_\epsilon}<\epsilon$$

Similarly for $n \le k$, $$d(-k,-n)<\epsilon$$ by the symmetric property of the metric. This shows that the sequence $\{-n\}$ in $X_4$ is a cauchy sequence and it is clearly not convergent as $-\infty$ is not in $\Bbb{R}$.

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