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Mar 2, 2023 at 11:44 vote accept Mehmet Onat
Mar 2, 2023 at 10:12 comment added Mehmet Onat Sorry I didn't know that thank you
Mar 1, 2023 at 21:19 comment added KP Hart Yes. (It would be wiser to have these in just one comment; hit shift+enter to go to a new line inside a comment.)
Mar 1, 2023 at 20:49 comment added Mehmet Onat Consequently, $d\left( x,y\right) <\delta \Rightarrow d_{n}\left( x,y\right) <\varepsilon $.
Mar 1, 2023 at 20:49 comment added Mehmet Onat $\Rightarrow d_{n}\left( x,y\right) <\frac{2}{a_{n}}d\left( x,y\right) < \frac{2}{a_{n}}\delta \leq \frac{2}{a_{n}}\cdot \frac{a_{n}\varepsilon }{2} =\varepsilon\Rightarrow d_{n}\left( x,y\right) <\varepsilon $.
Mar 1, 2023 at 20:48 comment added Mehmet Onat $\Rightarrow d_{n}\left( x,y\right) <1\Rightarrow 1+d_{n}\left( x,y\right) <2 $. From this, we get $\frac{d_{n}\left( x,y\right) }{2}<\frac{d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }$. Hence $\Rightarrow \frac{ a_{n}d_{n}\left( x,y\right) }{2}<\frac{a_{n}d_{n}\left( x,y\right) }{ 1+d_{n}\left( x,y\right) }\leq d\left( x,y\right) $
Mar 1, 2023 at 20:48 comment added Mehmet Onat We have $\frac{a_{n}d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }\leq d\left( x,y\right) $. Suppose that $d\left( x,y\right) <\delta $. Then $\frac{a_{n}d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }<\delta \leq \frac{a_{n}}{2}\Rightarrow \frac{d_{n}\left( x,y\right) }{1+d_{n}\left( x,y\right) }<\frac{1}{2}$.
Mar 1, 2023 at 20:47 comment added Mehmet Onat Let's choose $\delta \leq \min \left\{ \frac{a_{n}}{2},\frac{a_{n}\varepsilon }{2}\right\} $.
Mar 1, 2023 at 20:47 comment added Mehmet Onat For the second direction, I understand that Given $\varepsilon >0$ ve $n\in \mathbb{N}$
Mar 1, 2023 at 14:49 history edited KP Hart CC BY-SA 4.0
added 83 characters in body
Mar 1, 2023 at 9:48 history answered KP Hart CC BY-SA 4.0