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Let $X$ the river metric on the positive plane $(R^+)^2$: $d((x,y),(u,v)) = |y-v|$ if $x=u$ and $|y| + |v| + |x-u|$ otherwise.

(the x-axis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first).

Let $v_x$ the vertical line by orizonatal quote $x$, then the subspace topology of $v_x\setminus$ {$(x,0)$} is the euclidean topology, but any open that containing a point of $r_0$ the the bottom horizontal line is non-separable (involving a no countable vertical segments) we call a such open a "b-open" .

THen any covering $\mathcal{U}$ of $X$ as order 2, i.e. there exixt almost 3 elemets by no-empty intersection:

Considering a b-open $U\in \mathcal{U}$, exist a point $x\in r_o \cap (Cl(U)\setminus U)$ (i.e. in its boundary in $r_0$) then there exixt another b-open $V\in \mathcal{U}$ containing $x$, and $U\cap V$ is a b-open. The sets $U$, $V$ and $U\cap V$ are also open in the euclidean topology and we can assume also connected and containing its own projection on $r_0$, then because $R^2$ has dimension 2 follow that exist a point $(x', y)\notin U\cup V$ such that any open set containing $(x', y) $ intersect $U\cap V$ (otherwise we can make a cover of order 2, then any refinement has order 2). Then the element $F\in \mathcal{U}$ containing $(x', y)$ (like a open interval in the $v_{x'}$) meet also $U\cap V$ :

IF NO,the $sup$-extrem of the quote that $U\cap V$ can reachedin in the $v_{x'}$ vertical is minor of $y$. Any (open ball) $B_\epsilon(x)$ with $x\in r_0$ is like a halph-square triangle then if this is included in $U\cap V$ it dont meet the follow open part $S\subset R^2$:

Considering the two halph line by base in $x',y$ at right by $45°$ pendence and $-45°$ at left (like the grafh graph of $y=|x|$ traslated from origin to $(x', y)$) and consider the superior part $S$ of plane $R^2$ these halph-line cut off, by these halph line included.

but then $U\cap V$ being union of (open) balls dont meet $S$, then exixt a (euclidean) open neighbord of $(x', y)$ that dont meet $U\cap V$ .
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Let $X$ the river metric on the positive plane $(R^+)^2$: $d((x,y),(u,v)) = |y-v|$ if $x=u$ and $|y| + |v| + |x-u|$ otherwise.

(the x-axis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first).

Let $v_x$ the vertical line by orizonatal quote $x$, then the subspace topology of $v_x\setminus$ {$(x,0)$} is the euclidean topology, but any open that containing a point of $r_0$ the the bottom horizontal line is non-separable (involving a no countable vertical segments) we call a such open a "b-open" .

THen any covering $\mathcal{U}$ of $X$ as order 2, i.e. there exixt almost 3 elemets by no-empty intersection:

Considering a b-open $U\in \mathcal{U}$, exist a point $x\in r_o \cap (Cl(U)\setminus U)$ (i.e. in its boundary in $r_0$) then there exixt another b-open $V\in \mathcal{U}$ containing $x$, and $U\cap V$ is a b-open. The sets $U$, $V$ and $U\cap V$ are also open in the euclidean topology and we can assume also connected and containing its own projection on $r_0$, then because $R^2$ has dimension 2 follow that exist a point $(x', y)\notin U\cup V$ such that any open set containing $(x', y) $ intersect $U\cap V$ (otherwise we can make a cover of order 2, then any refinement has order 2). Then the element $F\in \mathcal{U}$ containing $(x', y)$ (like a open interval in the $v_{x'}$) meet also $U\cap V$ (if no,the :

IF NO,the $sup$-extrem of the quote that $U\cap V$ can reachedin in the $v_{x'}$ vertical is minor of $y$ but y$. Any (open ball) $B_\epsilon(x)$ with $x\in r_0$ is like a halph-square triangle then this happen also if we move this is included in $x'$ U\cap V$ it dont meet the follow open part $S\subset R^2$:

Considering the two halph line by base in a enought little horizontal interval$x',y$ at right by $45°$ pendence and $-45°$ at left (like the grafh of $y=|x|$ traslated from origin to $(x', y)$ and consider the superior part $S$ of plane $R^2$ these halph-line cut off, by these halph line included.

but then $U\cap V$ being union of (open) balls dont meet $S$, then exixt a (euclidean) open neighbord of $(x', y)$ that dont meet $U\cap V$ ) .
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Let $X$ the river metric on the positive plane $(R^+)^2$: $d((x,y),(u,v)) = |y-v|$ if $x=u$ and $|y| + |v| + |x-u|$ otherwise.

(the x-axis is the river, there are othogonal paths in which we can move "as in R", while otherwise we have to go via the river first).

Let $v_x$ the vertical line by orizonatal quote $x$, then the subspace topology of $v_x\setminus$ {$(x,0)$} is the euclidean topology, but any open that containing a point of $r_0$ the the bottom horizontal line is non-separable (involving a no countable vertical segments) we call a such open a "b-open" .

THen any covering $\mathcal{U}$ of $X$ as order 2, i.e. there exixt almost 3 elemets by no-empty intersection:

Considering a b-open $U\in \mathcal{U}$, exist a point $x\in r_o \cap (Cl(U)\setminus U)$ (i.e. in its boundary in $r_0$) then there exixt another b-open $V\in \mathcal{U}$ containing $x$, and $U\cap V$ is a b-open. The sets $U$, $V$ and $U\cap V$ are also open in the euclidean topology and we can assume also connected and containing its own projection on $r_0$, then because $R^2$ has dimension 2 follow that exist a point $(x', y)\notin U\cup V$ such that any open set containing $(x', y) $ intersect $U\cap V$ (otherwise we can make a cover of order 2, then any refinement has order 2). Then the element $F\in \mathcal{U}$ containing $(x', y)$ (like a open interval in the $v_{x'}$) meet also $U\cap V$ (if no,the $sup$-extrem of the quote that $U\cap V$ can reachedin in the $v_{x'}$ vertical is minor of $y$ but then this happen also if we move $x'$ in a enought little horizontal interval, then exixt a (euclidean) open neighbord of $(x', y)$ that dont meet $U\cap V$) .