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Let $d_{1}$ and $d_{2}$ be two metric space's to make up a set M;

subpart. a); Show that D is given with D(x,y)=$max\{d_{1}(x,y),d_{2}(x,y)\}$ also is a metric space on M;

a) When D is a metric space on the set M. I must find out if is positiv, symmetric and triangle inequality.

\textbf{Positiv:} When I know $d_{1}$ and $d_{2}$ both are metric spaces on the set M then it is fulfilling they $d_{1},d_{2}\leq 0$ and then D(x,y)=max $\{d_{1}(x,y),d_{2}(x,y)\} \geq 0$ I can conclude D is positiv.

\textbf{Symmetri:} D is symmetric and it's apply
D(x,y)=D(y,x) When D is given I know it must be equal to max of $d_{1}$ og $d_{2}$, and then D(x,y)=$d_{1}(x,y)$ or D(x,y)=$d_{2}(x,y)$ and those are both metrics and for them its apply $d_{1}(x,y)=d_{1}(y,x)$ and $d_{2}(x,y)=d_{2}(y,x)$ and then it must also apply D(x,y)=D(y,x) whatever D(x,y)=$d_{1}(x,y)$ or D(x,y)=$d_{2}(x,y)$ so I can conclude symmetric.

\textbf{Triangle inquality: I am tjekking if D is an triangle inequality $D(x,y)\leq D(x,z)+ D(z,y)$. Since it is previously true that D is given by max of $d_{1}$ and $d_{2}$, and thus D(x,y)=$d_{1}(x,y)$ or D(x,y)=$d_{2}(x,y)$ and since these are both metrics, it applies that $d_{1}(x,y)\leq d_{1}(x,z)+ d_{1}(z,y)$ and $d_{2}(x,y)\leq d_{2}(x,z)+ d_{2}(z,y)$ and thus since it applies that $D(x,y)\leq D(x,z)+D(z,y)$ no matter what D(x,y)=$d_{1}(x,y)$ eller D(x,y)=$d_{2}(x,y)$

subpart. b); Let a in M and r>0. Let$K_{1}(a,r), K_{2}(a,r) and K_{D}(a,r)$ describe the balls around a in M with the radius r, with regarding to the three metric spaces. ; Show that for 0<r and 0 <s it applies:

$$K_{D}(a,min\{r,s\} \subset K_{1}(a,r)\cap K_{2}(a,s)\subset K_{D}(a,max\{r,s\})$$

Let $K_{D}(a,min\{r,s\})\subseteq$ then i know that $min\{r,s\}\leq r\cap s \leq max\{r,s\}$ must happen. Then it must apply r<s, and $r\leq r\leq s$, when r>s is $s\leq s\leq e$ and when r=s then it will stand $s \leq s\leq s$ or $r \leq r\leq r$. It is applying for all 0 <r,s for $min\{r,s\}\leq r\cap s\leq max\{r,s\}$ and I can conclude it's must apply k $$K_{D}(a,min\{r,s\})\subseteq K_{1}(a,r) \cap K_{2}(a,s) \subseteq K_{D}(a,max\{r,s\}$$

subpart. c); Let A $\subset$ M. Show if A is open with regarding to at least one the metric spaces $d_{1}$ and $d_{2}$ then it's also open regarding to D.;

subpart. d); Assume A $\subset$ M is compact with regarding to D. Show that it is compact with regarding to both $d_{1}$ and $d_{2}:$

I can't figure out subpart c and d and my curriculum at Copenhagen university have covered all up to Fourier series, metric spaces last week.

Oliver Rommelhoff Reichel
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1 Answers1

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Let $U$ be open w.r.t. $d_1$ an $x \in U$. Then there eixsts $r>0$ such that $d_1(x,y) <r$ implies $y \in U$. Now $D(x,y) <r$ implies $d_1(x,y) <r$ and hence $y \in U$. Thus $U$ is open w.r.t. $D$. Similar argument holds when $u$ is open w.r.t. $d_2$.

d) follows immediately from c): Suppose $(U_i)$ is an open cover of $M$ w.r.t. $d_1$. Then it also an open cover w.r.t. $D$ and hence there is a finite subcover. So $M$ is compact w.r.t. $d_1$. Similarly it is compact w.r.t. $d_2$.

Kavi Rama Murthy
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