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November 25th, 2018, 11:57 AM  #1 
Newbie Joined: Nov 2018 From: Canada Posts: 4 Thanks: 0  Adherence of subset  Kernel
Having Y, a subspace of X. How can we show that the adherence of Y can be expressed as : Adherence of Y = intersection of { Ker(f)  f element of X* , Y contained in Ker(f)} 
November 25th, 2018, 12:07 PM  #2 
Senior Member Joined: Oct 2009 Posts: 752 Thanks: 261 
Use HahnBanach

November 25th, 2018, 12:22 PM  #3 
Newbie Joined: Nov 2018 From: Canada Posts: 4 Thanks: 0 
Can you give more details please ? 
November 25th, 2018, 03:09 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics  I'm assuming you are given that $X,Y$ are Banach spaces yes? Then fix $x_0 \in X \setminus \text{cl}(Y)$ and define a subspace, $X_0 = Y \bigcup \text{span}(x_0)$ and a linear functional, $x_0^* \in X_0^*$ by \[x_0^*(x) = \frac{d(x,Y)}{d(x_0,Y)} \] where $d: X \to \mathbb{R}$ is the set distance, $d(x,S) = \sup\{\left \leftxy \right \right: y \in S\}$. Now, show that you can extend $x_0^*$ to a linear functional on the entire space and that this extension is a competitor in your intersection since $Y \in \text{ker}(x_0^*)$. 
November 25th, 2018, 03:19 PM  #5  
Newbie Joined: Nov 2018 From: Canada Posts: 4 Thanks: 0  Quote:
Actually, it doesn't say anywhere in the exercise that its Banach spaces....  

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adherence, aherence, kernel, normed, space, subset, vector 
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