I dont know how to show the functions of this type if it is continuous or not in topological space. I know the method is to use one of two properties: the inverse image of eah open is open, and the inverse image of each closed is closed but i dont know how to use them. i will be thaked if someone help ed me with the steps if f:(R,U) −> (R,U) defined by f(x) = { (((1/3) x +1 , x>3)),(((1/2) (x+5) , x≦3)) :} Is f continuous?


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Question Number 79081 by arkanmath7@gmail.com last updated on 22/Jan/20

$I d o n t k n o w h o w t o s h o w t h e f u n c t i o n s o f$ $t h i s t y p e i f i t i s c o n t i n u o u s o r n o t i n t o p o l o g i c a l s p a c e .$ $I k n o w t h e m e t h o d i s t o u s e o n e o f t w o p r o p e r t i e s :$ $t h e i n v e r s e i m a g e o f e a h o p e n i s o p e n,$ $a n d t h e i n v e r s e i m a g e o f e a c h c l o s e d i s c l o s e d$ $b u t i d o n t k n o w h o w t o u s e t h e m .$ $i w i l l b e t h a k e d i f s o m e o n e h e l p e d m e w i t h t h e s t e p s$ $i f f : (R, U) - > (R, U) d e f i n e d b y$ $f (x) = {\begin{cases} \frac{1}{3} x + 1, x > 3 \\ \frac{1}{2} (x + 5), x ≦ 3 \end{cases}$ $I s f c o n t i n u o u s ?$
	Answered by mind is power last updated on 22/Jan/20

	$S i n c e I R i s m e t r i c a l w e c a n u s e s e q u e n c e s e$ $m e t r i c s p a c e \Rightarrow s e p a r a b a l \Rightarrow u n i c i t e o f l i m i t e s$ $U_{n} = 3 - \frac{1}{n} \to 3, V_{n} = 3 + \frac{1}{n} \to 3, w i t h e n \in N^{*}$ $U_{n} ⩽ 3, V_{n} > 3$ $f (U_{n}) = \frac{1}{2} (3 - \frac{1}{n} + 5) = 4 - \frac{1}{2 n} \to 4$ $f (V_{n}) = \frac{1}{3} (3 + \frac{1}{n}) + 1 = 2 + \frac{1}{3 n} \to 4$ $You can't use 'macro parameter character #' in math mode$
	Commented by arkanmath7@gmail.com last updated on 22/Jan/20


	Commented by arkanmath7@gmail.com last updated on 22/Jan/20

	$I w r o t e i s i t c o n t ? b u t t h e q u s t n i n t e s t s a y s s h o w i t i s c o n t .$
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