Show that $d$ is a metric for $X$ then , $d'(x,y) =\frac{d(x,y)}{1+d(x,y)}$ is a bounded metric space that gives the topology of $X$.

In dbfin.com the solution reads as follows :

Now, we show that d′ induces the same topology as d . Since $f$ and $f^{−1}(y)=\frac{y}{1−y}:[0,1)→R^+$ are continuous, $d′=f∘d$ and $d=f^{−1}∘d′$ , $d′$ is continuous in the $d$ -topology, and $d$ is continuous in the $d′$ -topology, implying that the topologies are the same .

Which topologies they are talking about? I think they talk about the coarsest topologies on $X \times X$ such that $d : X \times X \to \mathbb R$ and $d' : X \times X \to \mathbb R$ are continuous.

Can anyone please correct me if I have gone wrong anywhere?

  • 1
    $\begingroup$ Both $d$ and $d'$ induce a metric topology on $X$. $\endgroup$ – Joe Johnson 126 Aug 16 '18 at 13:00
  • $\begingroup$ metric topology means?@JoeJohnson126 $\endgroup$ – cmi Aug 16 '18 at 13:11
  • $\begingroup$ Are the coarsest topology on $X \times X$ such that the function $d : X \times X \to R$ is continuous and the metric topology on $X$ same?@JoeJohnson126 $\endgroup$ – cmi Aug 16 '18 at 13:46

A metric $d$ on a set $X$ induces a topology on $X$ (the metric topology generated by $d$) by using open balls as a local base at each point.

In other words, the topology consists of all arbitrary unions of sets of the form $B_d(x,\epsilon) \equiv \{y\in X\mid d(x,y)<\epsilon\}$.

Different metrics need not generate the same topology, but some do.

A simple example is that the discrete topology on $X$ (the topology is the set $\mathscr{P}(X)$ of all subsets of $X$) is generated by the metric $d_{disc}$ where $d_{disc}(x,y)=1$ if $y\neq x$ and $0$ if $y=x$; in this case, each point is an open ball, so every subset of $X$ is open in this metric topology.

  • $\begingroup$ You may have not read my question . Please go through my answer. Your answer is too trivial. I just want to know an information. I have problem in understanding which topologies they are talking about?@MPW $\endgroup$ – cmi Aug 16 '18 at 13:41
  • $\begingroup$ I knew what is metric topology but I ask to confirm. Please read the question once again.@MPW $\endgroup$ – cmi Aug 16 '18 at 13:42

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