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Let $T$ be an element of $B(X,X)$ (the bounded linear operators from X to itself), and let $W$ be a subset of $X$. Show that $T(\overline{W}) \subset \overline{T(W)]}$. Furthermore, if $T$ has bounded inverse, then $T(\overline{W}) = \overline{[T(W)]}$.

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Try \bar{W} ($\bar{W}$) or \overline{W} ($\overline{W}$). –  Calle May 10 '11 at 22:33
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Use \overline{...}. Concerning your question. You don't say what $X$ is but I guess it's a Banach space. Note that this has nothing to do with $X$ being a Banach space and linearity of $T$ (except for the trivial fact bounded = continuous). It's just an equivalent reformulation of continuity of $T$ (and of $T^{-1}$) between topological spaces. –  t.b. May 10 '11 at 22:34

1 Answer 1

Take a point $x$ in the closure of $W$ and a sequence in $W$ converging to $x$. The image of the sequence under $T$ is in $T(W)$ and it converges to $T(x)$ since $T$ is bounded, so $T(x)$ is in the closure of $T(W)$.

For the second bit, think about applying $T^{-1}$ to both sides of the inclusion.

[Maybe \overline is what you're looking for? $\overline{W}$]

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