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Let $X$ and $Y$ be normed vector spaces and $T$ be a linear operator from $X$ to $Y$ whose norm is $1$. If $U$ is an open set in $X$, is it true that $\text{closure}(T(U))-\text{closure}(T(U))=\text{closure}(T(U)-T(U))$?

Here, $A-B$ is a set of $(a-b)$'s where $a$ and $b$ are elements of $A$ and $B$ respectively.

I think that statement is false, but i cannot find the counterexample.

Andrew Uzzell
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tuko
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1 Answers1

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If $Z$ is a subspace of $Y$, then $Z-Z=Z$. Moreover, the closure of a subspace is a subspace. So $\text{cl}(Z-Z)=\text{cl}(Z)=\text{cl}(Z)-\text{cl}(Z)$.

Seirios
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