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The following statement is true:

Assume that $X$ and $Y$ are topological vector spaces where $Y$ is finite-dimensional Hausdorff, if $A:X\rightarrow Y$ is a continuous surjective linear map then $A$ is an open map.

With the same setting, I am looking for a discontinuous $A$ which is not an open map, can anyone suggest any of such examples?

Moreover, what about any example that satisfies

1) $Y$ finite-dimensional but not Hausdorff, $A$ continuous surjective linear map but not open.

2) $Y$ finite-dimensional but not Hausdorff, $A$ discontinuous surjective linear map but not open.

3) Similar cases apply on infintie dimensional $Y$.

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With the same setting, I am looking for a discontinuous $A$ which is not an open map, can anyone suggest any of such examples?

That doesn't exist. If $A\colon X \to Y$ is a surjective linear map, where $X,Y$ are topological vector spaces and $Y$ is finite-dimensional and Hausdorff, then $A$ is open, regardless of continuity. One way to see that is to note that if we replace the topology on $X$ by a finer one, it becomes easier for a map to be continuous, but harder to be open. If we endow $X$ with the finest vector space topology on $X$, then $A$ is continuous, hence open by what you know. But every open set in the topology on $X$ we started with is open in the finest vector space topology, hence its image under $A$ is open. Another way to see it is to consider any section of $A$, that is a linear map $S \colon Y \to X$ such that $A\circ S = \operatorname{id}_Y$. Since $Y$ is finite-dimensional Hausdorff, $S$ is continuous. But for an open set $U\subset X$ we have $A(U) = A(U + \ker A) = S^{-1}(U + \ker A)$, which is open by the continuity of $S$.

Coming to the examples with non-Hausdorff $Y$, let $E$ denote $\mathbb{R}$ endowed with the standard topology, and let $F$ denote $\mathbb{R}$ endowed with the indiscrete topology. Then $E$ and $F$ are finite-dimensional topological vector spaces, hence so is $E\times F$.

Ad 1), we can take $\operatorname{id} \colon E \to F$. It's clearly continuous and surjective, but of course not open.

Ad 2), we can take $\operatorname{id}\times \operatorname{id} \colon F\times E \to E\times F$. Again it's clearly surjective, but the first component is not continuous, so the whole map is not continuous, and the second component is not open, which makes the entire map not open.

For the finite-dimensional case, all examples are somewhat similar to these examples, since every finite-dimensional topological vector space over $\mathbb{R}$ is topologically isomorphic to $E^m\times F^n$ for some $m,n\in \mathbb{N}$.

For the infinite-dimensional case, we can easily modify the examples (take a product of a Hausdorff and an indiscrete space).

Somewhat more interesting may be the following example: Let

$$c_{00} = \bigl\{ f \colon \mathbb{N}\to \mathbb{R} \mid \bigl(\exists k\bigr)\bigl(n \geqslant k \implies f(n) = 0\bigr)\bigr\}$$

be the space of all real sequences with finite support. Endow it with the subspace topology induced by your favourite $\ell^p(\mathbb{N})$, and consider the map $A \colon c_{00} \to c_{00}$ given by $(Af)(n) = 2^{-n}\cdot f(n)$. Then $A$ is continuous and bijective, but not open. If we endow the codomain with a coarser non-Hausdorff topology, the map remains continuous and non-open.

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  • $\begingroup$ Is the $S$ defined in the argument of the first statement really continuous? $S:Y\rightarrow X$, $Y$ is finite-dimensional Hausdorff, but $X$ here is not necessary to be Hausdorff, as what I know is the following theorem: A linear map between Hausdorff TVS where the domain is finite dimensional, then the linear map is continuous. Of course, the very beginning argument that considering the finest topology and so on works well. $\endgroup$ – user284331 Dec 17 '15 at 5:25
  • $\begingroup$ On a finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ (I've tacitly assumed that we consider only those scalar fields, I don't know how it is for arbitrary topological fields, but whenever it's true that every linear map between Hausdorff TVSs with finite-dimensional domain is continuous, it holds), there is a unique Hausdorff vector space topology. Thus $Y$ is topologically isomorphic to $\mathbb{K}^n$ (endowed with the product topology), and the continuity of a linear map $L\colon \mathbb{K}^n \to X$ directly follows from the definition of a TVS, $\endgroup$ – Daniel Fischer Dec 17 '15 at 9:33
  • $\begingroup$ the continuity of scalar multiplication gives the continuity of $(c_1,\dotsc,c_n) \mapsto (c_1\cdot L(e_1),\dotsc, c_n\cdot L(e_n))$, and the continuity of addition gives the continuity of $(v_1,\dotsc, v_n) \mapsto v_1 + \dotsc + v_n$. The composition of these maps is $L$, which is thus seen to be continuous. Or you can note that the finest vector space topology on $X$ is Hausdorff, so $S$ is continuous if we endow $X$ with the finest TVS topology. But making the topology on the codomain coarser leaves continuous maps continuous (and can make more maps continuous). $\endgroup$ – Daniel Fischer Dec 17 '15 at 9:34

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