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  1. Continuous linear mappings between topological vector spaces preserve boundedness.

    I was wondering if it means that the inverse image of a bounded subset under a continuous linear mapping is still bounded?

    Conversely, must a mapping between two topological vector spaces, such that the inverse image of any bounded subset is still bounded, be continuous linear?

  2. A continuous linear operator maps bounded sets into bounded sets.

    Does it mean that the image of a bounded subset under a continuous linear mapping is still bounded?

    Conversely, must a mapping between two topological vector spaces that maps bounded sets to bounded sets be continuous linear?

Thanks and regards!

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  • $\begingroup$ The answer to the first question is clearly no, since the mapping can collapse the domain to the zero vector. $\endgroup$ Commented Feb 25, 2012 at 2:07
  • $\begingroup$ The answer to the second is yes. But the second is not true suppose for example a first space as a separable space and put ${x_{k}}$ a dense sequence then the function $f$ such that $f(x_{k})=1$ and zero case contrary, this function is neither continuous nor liner but naturally bounded. $\endgroup$
    – checkmath
    Commented Feb 25, 2012 at 2:37

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The answer to the first question is clearly no, since the mapping can collapse the domain to the zero vector. A function that simply interchanges two points has an inverse that takes bounded sets to bounded sets, but the function is neither continuous nor linear.

The second statement is precisely equivalent to the first, so it does indeed mean that the image of a bounded set under a continuous linear mapping is bounded. The answer to the final question is no, just as in the first part.

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  • $\begingroup$ +1 Thanks! (1) are you saying the two quotes from Wiki are wrong? (2) Why "The second statement is precisely equivalent to the first"? (My bad, I should keep my last comment. Could you undelete your response to my last comment? I am not able to undelete mine.) $\endgroup$
    – Tim
    Commented Feb 25, 2012 at 2:21
  • $\begingroup$ @Tim: No, they’re fine; they simply don’t say anything about the inverses of continuous linear maps. They’re equivalent because they mean exactly the same thing: if $B$ is bounded, then $f[B]$ is bounded. $\endgroup$ Commented Feb 25, 2012 at 2:24
  • $\begingroup$ Thanks! Generally when saying a mapping preserving some property of subsets, (1) doesn't it mean that the inverse image under the mapping of any codomain subset with the property must also have that property in the domain? For example, a continuous mapping preserves openness, a measurable mapping preserves measurability, ... (2) Or preserving subset property means the image of any domain subset with the property must have that property in the codomain? Like an open mapping for openness of subsets? $\endgroup$
    – Tim
    Commented Feb 25, 2012 at 2:29
  • $\begingroup$ @Tim: No, it just means that if something in the domain has the property, so does its image. A continuous mapping does not necessarily preserve openness; a map that preserves openness is one that sends open sets to open sets, and such a map is called an open map. A continuous map need not be open, and an open map need not be continuous. $\endgroup$ Commented Feb 25, 2012 at 2:33
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    $\begingroup$ @Tim: Unfortunately, that statement is a bit misleading. It would be better to say that morphisms respect the structure, meaning that their behavior is in some important way related to the structure. They don’t in general preserve all aspects of structure: when they do, they are isomorphisms in linear algebra and group theory and homeomorphisms in topology. In particular, a continuous function definitely does not necessarily preserve openness, though it does preserve other aspects of topological structure (e.g., convergence of sequences). $\endgroup$ Commented Feb 25, 2012 at 2:54

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