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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: $$\bigcap\{f(U_\alpha):\alpha\in\Lambda\}\subseteq f\left(\bigcap\{U_\alpha:\alpha \in \Lambda\}\right)$$

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  • $\begingroup$ BTW $\land$ is more usually as notation for a logical connective. Perhaps you were trying to write $\Lambda$? $\Lambda$ $\endgroup$ – Martin Sleziak Dec 6 '13 at 14:41
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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 $\alpha$?

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  • $\begingroup$ 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? $\endgroup$ – Randomstop Dec 6 '13 at 3:09
  • $\begingroup$ @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? $\endgroup$ – T. Eskin Dec 6 '13 at 3:11
  • $\begingroup$ The only relationship I know of occurs when $f(x_1)=f(x_2)$ then $x_1=x_2$ $\endgroup$ – Randomstop Dec 6 '13 at 3:15
  • $\begingroup$ @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 $x_{\beta}$? $\endgroup$ – T. Eskin Dec 6 '13 at 3:19
  • $\begingroup$ Oh! So were just saying that then every $x_\alpha=x_\beta...$ So then $x\in U_\alpha$ and $f(x)=y$ $\endgroup$ – Randomstop Dec 6 '13 at 3:21
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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?

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  • $\begingroup$ 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 $x_1=x_2$ $\endgroup$ – Randomstop Dec 6 '13 at 3:04

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