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$(X,d)$ be a metric space and define $p(x,y)=\min\{{1\over 2},d(x,y)\}$ $\mathfrak{T_1},\mathfrak{T}_2$ be the topology generated by $d,p$, I need to know relations between them.

Well, open sets of $\mathfrak{T}_1$ are smaller than the other one that I can see easily, so can I say $\mathfrak{T_1}\subseteq\mathfrak{T}_2$ ?

Myshkin
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  • See http://math.stackexchange.com/questions/156318/proof-that-every-metric-space-is-homeomorphic-to-a-bounded-metric-space and http://math.stackexchange.com/questions/275841/standard-bounded-metric-induces-same-topology – Chris Culter Aug 15 '13 at 06:18
  • @ChrisCulter Thank you very much for these links – Myshkin Aug 15 '13 at 06:37

3 Answers3

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You can say more, namely $\mathfrak T_1=\mathfrak T_2$.

Assume $U\in \mathfrak T_1$. Then for every $x\in U$ there is an open ball around $x$ contained in $U$, that is for some $r>0$, all $y$ with $d(y,x)<r$ belong to $U$. May may as well assume that $r<\frac12$ and in that case also all $y$ with $p(y,x)<r$ are in $U$. Hence $U\in\mathfrak T_2$.

The same idea that we may assume wlog. that $r<\frac12$ works just as well to show that $U\in\mathfrak T_2$ implies $U\in\mathfrak T_1$.

Hagen von Eitzen
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Open sets in a metric spaces are completely determined by all open balls of radius at most $r$, for any $r>0$. The reason is that any open ball in a metric space is the union of open balls of radius at most $r$. So, the two topologies you describe are equal.

Ittay Weiss
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Note that in order to show that two topologies are equal it suffices to prove that there are bases $\cal B_i$ of $\frak T_i$ such that every open set in $\frak T_1$ contains a basic open set from $\cal B_2$ and vice versa.

Hint: Show that $\left\{B_r(x)\mathrel{}\middle|\mathrel{} x\in X, r<\frac12\right\}$ is a basis for both topologies.

Asaf Karagila
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