Topolojik Denk Metrikler-I

Question

$(X,d_1),(X,d_2)$ metrik uzaylar olmak üzere

$$d_1\overset{T}{\sim} d_2$$

$$\Leftrightarrow$$

$$(\forall\epsilon>0)(\forall x\in X)(\exists\delta_1,\delta_2>0)[B^{d_1}(x,\delta_1)\subseteq B^{d_2}(x,\epsilon)\wedge B^{d_2}(x,\delta_2)\subseteq B^{d_1}(x,\epsilon)]$$

Tanım: $(X,d_1),(X,d_2)$ metrik uzaylar olmak üzere

$$d_1\overset{T}{\sim} d_2:\Leftrightarrow \tau_{d_1}=\tau_{d_2}$$

Not: $(X,d)$ metrik uzay ve $A\subseteq X$ olmak üzere

$$A,  \ d\text{-açık}:\Leftrightarrow (\forall a\in A)(\exists\epsilon >0)(B^d(a,\epsilon)\subseteq A) $$

$$\tau_d:=\{A|(A\subseteq X)(A,  \ d\text{-açık})\}$$

Answer 1

Gerek Kısmı: $d_1\overset{T}{\sim} d_2, \epsilon>0$ ve $x\in X$ olsun. 

$\left.\begin{array}{rr}d_1\overset{T}{\sim} d_2\Rightarrow \tau_{d_1}=\tau_{d_2} \\ \\ (\epsilon>0)(x\in X)\Rightarrow B^{d_1}(x,\epsilon)\in \tau_{d_1} \end{array}\right\}\Rightarrow x\in B^{d_1}(x,\epsilon)\in \tau_{d_2}$

 

$\Rightarrow (\exists \delta>0)(B^{d_2}(x,\delta)\subseteq B^{d_1}(x,\epsilon)).$

 

Benzer şekilde 

$\left.\begin{array}{rr}d_1\overset{T}{\sim} d_2\Rightarrow \tau_{d_1}=\tau_{d_2} \\ \\ (\epsilon>0)(x\in X)\Rightarrow B^{d_2}(x,\epsilon)\in \tau_{d_2} \end{array}\right\}\Rightarrow x\in B^{d_2}(x,\epsilon)\in \tau_{d_1}$

 

$\Rightarrow (\exists \delta>0)(B^{d_1}(x,\delta)\subseteq B^{d_2}(x,\epsilon)).$

$$-----------------------------------$$

Yeter Kısmı: $d_1\overset{T}{\sim} d_2$ olduğunu göstermek için $\tau_{d_1}=\tau_{d_2}$ olduğunu göstermeliyiz. Bunun için de $\tau_{d_1}\subseteq \tau_{d_2}$ ve $\tau_{d_2}\subseteq \tau_{d_1}$ olduğunu göstermek gerekli ve yeterli olacaktır.

$\left.\begin{array}{rr} A\in \tau_{d_1}\Rightarrow (\forall a\in A)(\exists \epsilon>0)(B^{d_1}(a,\epsilon)\subseteq A) \\ \\ \text{Hipotez} \end{array}\right\}\Rightarrow$

$\Rightarrow (\forall a\in A)(\exists \delta>0)(B^{d_2}(a,\delta)\subseteq B^{d_1}(a,\epsilon)\subseteq A)$

$\Rightarrow A\in \tau_{d_2}$

O halde $\tau_{d_1}\subseteq \tau_{d_2}\ldots (1)$

 

$d_1\overset{T}{\sim} d_2$ ve $A\in \tau_{d_2}$ olsun.

$\left.\begin{array}{rr} A\in \tau_{d_2}\Rightarrow (\forall a\in A)(\exists \epsilon>0)(B^{d_2}(a,\epsilon)\subseteq A) \\ \\ \text{Hipotez} \end{array}\right\}\Rightarrow$

$\Rightarrow \forall a\in A)(\exists \delta>0)(B^{d_1}(a,\delta)\subseteq B^{d_2}(a,\epsilon)\subseteq A)$

$\Rightarrow A\in \tau_{d_1}$

O halde $\tau_{d_2}\subseteq \tau_{d_1}\ldots (2)$

$$(1),(2)\Rightarrow \tau_{d_1}=\tau_{d_2}\Rightarrow d_1\overset{T}{\sim} d_2.$$