About $\rho(x, y)= \begin{cases}\mathrm{d}(x, y) & \mathrm{d}(x, y)<1 \\ 1 & \mathrm{~d}(x, y) \geq 1\end{cases}$ in a metric space

let $(X,d)$ be a metric space now define $\rho(x, y)= \begin{cases}\mathrm{d}(x, y) & \mathrm{d}(x, y)<1 \\ 1 & \mathrm{~d}(x, y) \geq 1\end{cases}$ now which of following options is false ?

1. $(X,d)$ and $(X,\rho)$ have same open sets .

2. $(X,d)$ and $(X,\rho)$ have same compact sets .

3. $(X,d)$ and $(X,\rho)$ have same connected sets .

4. $(X,d)$ and $(X,\rho)$ have same bounded sets .

Two metrics $d,d'$ are called equivalent if there exist constants $c,C>0$ such that for all $x,y \in X$:

$$ c d(x,y) \le d'(x,y) \le C d(x,y)$$

first of all $X$ has infinite element because for finite case every metric on $X$ is equal to discrete metric .so let $X= \mathbb{R}$ and $ d(x,y)=|x-y|$ so $\mathbb Q$ is not bounded in $(X,d)$ but in $(X,\rho)$ is bounded so "4" is false . how we can show another options are true ?

Solution 1:

The metrics $d$ and $\rho$ induce the same open sets, and therefore the first three options are true.

They induce the same open sets because, if $A$ is an open set with respect to the metric $d$, then for each $x\in A$, there is some $r(x)>0$ such that the open $d$-ball centered at $x$ with radius $r(x)$ is contained in $A$. You can assume, without loss of generality, that $r(x)<1$, and then the open $d$-ball centered at $x$ with radius $r(x)$ is the open $\rho$-ball centered at $x$ with radius $r(x)$. Since $A$ is the union of all these balls, $A$ is open with respect to the metric $\rho$.

And if $A$ is an open set with respect to the metric $\rho$, then for each $x\in A$, there is some $r(x)>0$ such that the open $\rho$-ball centered at $x$ with radius $r(x)$ is contained in $A$. Again, you can assume, without loss of generality, that $r(x)<1$, and then the open $\rho$-ball centered at $x$ with radius $r(x)$ is the open $d$-ball centered at $x$ with radius $r(x)$. Since $A$ is the union of all these balls, $A$ is open with respect to the metric $d$.

Solution 2:

The Last option is false. You showed that $\mathbb{Q}$ is bounded in $(\mathbb{R},\rho)$. Which suffices. Even more broadly, the entire space $(\mathbb{R},\rho)$ is bounded. where as $\mathbb{R}$ with usual topology is not bounded.

Two metrics are equivalent if and only if for any open set $U$ in $(X,d)$ is open in $(X,\rho)$. . This is an equivalent formulation . The condition you state in your question would be called Strongly Equivalent. If that holds then even the bounded sets will be the same . (Because you can shrink and expand the balls by usage of those finite $c,C$ and get boundedness.

$d$ and $\rho$ are equivalent and hence you will have a homeomorphism from $(X,d)\to (X,\rho)$ and all topological properties will be same.

To prove that the metric topology generated by $d$ is same as that generated by $\rho$ all you have to do is:-

1.Take an open ball $(X,d)$ and show that this ball is open in $(X,\rho)$. 2. Take an open ball in $(X,\rho)$ and show that this ball is open in $(X,d)$

However $(X,d)$ may be unbounded but $(X,\rho)$ is always bounded.

Alternatively you can show that the identity map $i:(X,d)\to (X,\rho)$ such that $i(x)=x$ is continuous. to show that this is a homeomorphism and hence these two spaces will have same open , closed, connected and compact sets. Well for any $0\leq\epsilon<1$ we have $|i(x)-i(y)|<\epsilon$ for any $|x-y|<\epsilon$ . And for $\epsilon\geq 1$.

$|i(x)-i(y)|< \epsilon$ for any $x,y\in (X,\rho)$.

And the inverse is also the identity map which is also continuous. That is the identity map is itself an Homeomorphism.

For the Proof of open sets are same:-

Let $U$ be an open set in $(X,d)$. Thus for any $y\in U$ . You have a ball $B_{d}(y,\epsilon)$ such that $B_{d}(y,\epsilon)\subset U$ . Thus you choose $n\in\mathbb{N}$ such that $\frac{\epsilon}{n}<1$. And then you have that $B_{d}(y,\frac{\epsilon}{n})\subset B_{d}(y,\epsilon)\subset U$.

But $B_{d}(y,\frac{\epsilon}{n})=B_{\rho}(y,\frac{\epsilon}{n})$.

So for all $y\in U$ . You have $B_{\rho}(y,\frac{\epsilon}{n})\subset U$. Thus $U$ is open in $(X,\rho)$ as $\displaystyle U=\bigcup_{y\in U}B_{\rho}(y,\frac{\epsilon}{n})$ and union of open sets is open.

Conversely let $V$ be open in $(X,\rho)$. Then for $y\in V$ we have $B_{\rho}(y,\epsilon)\subset V$. If $0<\epsilon<1$ then $B_{\rho}(y,\epsilon)=B_{d}(y,\epsilon)\subset V $ and hence $V$ is open in $(X,d)$. As $V=\bigcup_{y\in V} B_{d}(y,\epsilon)$.

If $\epsilon\geq 1$ then $ B_{\rho}(y,\epsilon)=X$. And $X$ is open in $(X,d)$ by definition.

Thus in any case $(X,\tau_{d})=(X,\tau_{\rho})$.

$\tau_{d}$ refers to the topology induced by the metric $d$.