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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-i

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Question Number 79081 by arkanmath7@gmail.com last updated on 22/Jan/20
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?
$${I}\:{dont}\:{know}\:{how}\:{to}\:{show}\:{the}\:{functions}\:{of}\: \\ $$$${this}\:{type}\:{if}\:{it}\:{is}\:{continuous}\:{or}\:{not}\:{in}\:{topological}\:{space}. \\ $$$${I}\:{know}\:{the}\:{method}\:{is}\:{to}\:{use}\:{one}\:{of}\:{two}\:{properties}: \\ $$$${the}\:{inverse}\:{image}\:{of}\:{eah}\:{open}\:{is}\:{open}, \\ $$$${and}\:{the}\:{inverse}\:{image}\:{of}\:{each}\:{closed}\:{is}\:{closed} \\ $$$${but}\:{i}\:{dont}\:{know}\:{how}\:{to}\:{use}\:{them}. \\ $$$${i}\:{will}\:{be}\:{thaked}\:{if}\:{someone}\:{help}\:{ed}\:{me}\:{with}\:{the}\:{steps} \\ $$$$ \\ $$$${if}\:{f}:\left(\mathbb{R},{U}\right)\:−>\:\left(\mathbb{R},{U}\right)\:{defined}\:{by} \\ $$$$ \\ $$$${f}\left({x}\right)\:=\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{3}}\:{x}\:+\mathrm{1}\:\:\:,\:{x}>\mathrm{3}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\left({x}+\mathrm{5}\right)\:\:,\:{x}\leqq\mathrm{3}}\end{cases} \\ $$$${Is}\:{f}\:{continuous}? \\ $$
Answered by mind is power last updated on 22/Jan/20
Since IR is metrical we can use sequencese metric space ⇒ separabal ⇒unicite of limites U_n =3−(1/n)→3,V_n =3+(1/n)→3, withe n∈N^∗ U_n ≤3,V_n >3 f(U_n )=(1/2)(3−(1/n)+5)=4−(1/(2n))→4 f(V_n )=(1/3)(3+(1/n))+1=2+(1/(3n))→4 since lim_(n→∞) f(U_n )#lim_(n→∞) f(V_n )⇒f is note vontinus over 3
$${Since}\:{IR}\:{is}\:{metrical}\:{we}\:{can}\:{use}\:{sequencese} \\ $$$${metric}\:{space}\:\Rightarrow\:{separabal}\:\Rightarrow{unicite}\:{of}\:{limites} \\ $$$${U}_{{n}} =\mathrm{3}−\frac{\mathrm{1}}{{n}}\rightarrow\mathrm{3},{V}_{{n}} =\mathrm{3}+\frac{\mathrm{1}}{{n}}\rightarrow\mathrm{3},\:{withe}\:{n}\in\mathbb{N}^{\ast} \\ $$$${U}_{{n}} \leqslant\mathrm{3},{V}_{{n}} >\mathrm{3} \\ $$$${f}\left({U}_{{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}−\frac{\mathrm{1}}{{n}}+\mathrm{5}\right)=\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}{n}}\rightarrow\mathrm{4} \\ $$$${f}\left({V}_{{n}} \right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{3}+\frac{\mathrm{1}}{{n}}\right)+\mathrm{1}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{3}{n}}\rightarrow\mathrm{4} \\ $$$${since}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({U}_{{n}} \right)#\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({V}_{{n}} \right)\Rightarrow{f}\:{is}\:{note}\:{vontinus}\:{over}\:\mathrm{3} \\ $$
Commented by arkanmath7@gmail.com last updated on 22/Jan/20
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Commented by arkanmath7@gmail.com last updated on 22/Jan/20
I wrote is it cont? but the qustn in test says show it is cont.
$${I}\:{wrote}\:{is}\:{it}\:{cont}?\:{but}\:{the}\:{qustn}\:{in}\:{test}\:{says}\:{show}\:{it}\:{is}\:{cont}. \\ $$

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