Weak Convergence of A in l^2: Does it Preserve Closedness?

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The discussion centers on the weak convergence of the sequence x_n = (1-1/n)e_n in the Hilbert space l^2. It establishes that the set A = {x_n} is closed in l^2, yet its image under the mapping x |--> ||x||^2 is not closed in R, as it fails to include the accumulation point 1. The poster questions whether the mapping remains closed under the weak topology on R, noting that in finite-dimensional spaces, the weak topology aligns with the norm topology.

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HMY
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Take the sequence x_n = (1-1/n)e_n in l^2
Consider the map l^2 to R given by x |--> ||x||^2

The set A = {x_n} in l ^2 is closed & its image is not closed in R
under the norm topology (it doesn't contain its accumulation point 1).
So ultimately the above map is not closed.

What I'm not sure about is whether the map is closed if I consider the
weak topology on R instead of the norm topology? I think my misunderstandings are arising from the weak convergence.
 
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Because R is finite-dimensional its weak topology coincides with the norm topology.

(I realize this post is really old, but I was bored and decided to browse the forum.)
 

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