MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a space, and $d_1$ and $d_2$ be two metrics on $X$.

Define $S(x,y)= ${$\Sigma_2^l Min${$d_1(x_{k-1},x_k),d_2(x_{k-1},x_k)$}$:x_1=x, x_l=y, l finite $} $x$ and $y$ are two points in $X$ The set considers all finite stepped routes from $x$ to $y$.

Define $d_1\wedge d_2(x,y)=inf${$S(x,y)$} as the meet of $d_1$ and $d_2$.

Define $d_1\vee d_2(x,y)=Max${$d_1(x,y),d_2(x,y)$} as the join of $d_1$ and $d_2$.

Prove that the set $D$ of all metrics on $X$ is a lattice under the standard definition of order($d_1>d_2$ if $d_1$ induces a finer topology than $d_2$)with this definition of meet and join.

The order relations can be written as $d_1>d_2$ if $\forall x,\epsilon, \exists {\epsilon}' $such that, $d_1(x,y)<{\epsilon}' \implies d_2(x,y)<\epsilon$

Essentially, the problem is to show that the defined meet is a metric and is indeed a meet; that boils down to finding a $non-zero$ lower bound to $S(x,y)$ in terms of $d_1(x,y)$ and $d_2(x,y)$

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.