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Trying to prove that if f is one-to-one, then $$f\left(\bigcap\{U_\alpha:\alpha \in \Lambda\}\right)=\bigcap\{f(U_\alpha):\alpha\in\Lambda\}$$

I am able to prove that: $$f\left(\bigcap\{U_\alpha:\alpha \in \Lambda\}\right)\subseteq\bigcap\{f(U_\alpha):\alpha\in\Lambda\}$$

However, I do not really know where to begin the proof for: $$\bigcap\{f(U_\alpha):\alpha\in\Lambda\}\subseteq f\left(\bigcap\{U_\alpha:\alpha \in \Lambda\}\right)$$

Brian M. Scott
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Randomstop
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2 Answers2

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Hint:

Take $y\in\bigcap f(U_{\alpha})$. Then for all $\alpha$ there exists $x_{\alpha}\in U_{\alpha}$ with $f(x_{\alpha})=y$. But since $f$ is one-to-one then what does it tell you about these specific $x_{\alpha}$'s? Can you express $y$ as $f(x)$ where $x\in U_{\alpha}$ for all $\alpha$?

T. Eskin
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  • Because f is one-to-one then $x_\alpha = y$ because each element in the range is associated with only one element of the domain? – Randomstop Dec 06 '13 at 03:09
  • @Randomstop: Since $x_{\alpha}$ is in the domain and $y$ in the range, we can't necessarily say that $x_{\alpha}=y$. But does the injectivity of $f$ imply something about the relationship of each $x_{\alpha}$ to each other? – T. Eskin Dec 06 '13 at 03:11
  • The only relationship I know of occurs when $f(x_1)=f(x_2)$ then $x_1=x_2$ – Randomstop Dec 06 '13 at 03:15
  • @Randomstop. Exactly. So in this scenario you know that $f(x_{\alpha})=y$ for all $\alpha$. So, if you take any indices $\alpha,\beta$, what do you know of $f(x_{\alpha})$ and $f(x_{\beta})$? And what does that tell you about the relationship of $x_{\alpha}$ and $x_{\beta}$? – T. Eskin Dec 06 '13 at 03:19
  • Oh! So were just saying that then every $x_\alpha=x_\beta...$ So then $x\in U_\alpha$ and $f(x)=y$ – Randomstop Dec 06 '13 at 03:21
  • @Randomstop. Yep, you got it. – T. Eskin Dec 06 '13 at 03:21
  • So then $y\in f(U_\alpha)$ ? – Randomstop Dec 06 '13 at 03:24
  • @Randomstop. Ideally you would want that $y\in f(\bigcap_{\alpha\in\Lambda} U_{\alpha})$. But you already know that $y=f(x)$ where $x$ is a member of all $U_{\alpha}$ (by considering $x=x_{\alpha}=x_{\beta}=..$). So you're practically almost there. – T. Eskin Dec 06 '13 at 03:26
  • Because $x\in \alpha$ for all $\alpha$, then $x\in\bigcap { U_\alpha : \alpha \in \land }$ Then we can take the image of both sides or whatever thats called so its like $f(x) \in f(\bigcap...)$. Then from there say $y \in f(\bigcap ...)$? – Randomstop Dec 06 '13 at 03:30
  • @Randomstop. Precisely. Except for the first line you probably meant $x\in U_{\alpha}$ for all $\alpha$ instead of $x\in \alpha$ for all $\alpha$. Otherwise, what you wrote is exactly what you wanted to conclude. – T. Eskin Dec 06 '13 at 03:36
  • @Randomstop. You're welcome. By the way. If you found this answer useful, and also in case for future answers, you can always up vote them by clicking the "arrow up" next to it and click the "accept" mark below the score meter. – T. Eskin Dec 08 '13 at 16:02
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There's only one way to prove that $A\subseteq B$ for two sets $A$ and $B$. You say “Let $x\in A$ be given.” Then you show that $x\in B$ also. So here you say “Let $x\in\bigcap\{f(U_\alpha):\alpha\in\land\}$.” Since you know that $x\in\bigcap\{f(U_\alpha):\alpha\in\land\}$, what do you know about $x$? What property or properties must it have?

You want to show that $x\in f\left(\bigcap\{U_\alpha:\alpha \in \land\}\right)$ also. What property would $x$ have to have for this to hold? What are you trying to prove?

MJD
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  • I know in order to show $x\in f\left(\bigcap{U_\alpha:\alpha \in \land}\right)$ the function has to be one-to-one. A function is one-to-one where $f(x_1)=f(x_2)$ and $x_1=x_2$ – Randomstop Dec 06 '13 at 03:04