## [SOLVED] Inverse Image of a Compact Set -- Bounded?

Problem:

Let f : X → Y be a continuous function, K ⊂ Y - compact set. Is it true that f$$^{-1}$$(K)– the inverse image of a compact set– is bounded? Prove or provide counterexample.

Questions Generated:

1. Why does compactness matter? (I know it does.)
2. Does knowing that the inverse of a continuous function sends closed sets to closed sets (and open to open) matter?

Solution Ideas:

Well, I just feel like this is false and has a counterexample. But I'm not sure. I was thinking that somehow I can show the inverse sends compact sets to compact sets, and that compact sets are closed and bounded. But this is only true in $$\mathbb{R}$$$$^{n}$$. So I'm not really sure.
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 Recognitions: Homework Help Science Advisor Firstly I should point out that a continuous map need NOT pull back compact sets to compact sets; maps that do this are called proper. Now as to the question at hand, try to think about how nice continuous functions on general metric spaces are. How nice they are depends on the metric, no? So try to give X some metric that doesn't restrict the set of continuous maps on X, like the discrete metric.
 If I choose X as the discrete metric, won't any function sending any set to X then be discrete, as X is? Because then it would be bounded, because all sets in the discrete metric are bounded by 1. So I'm not exactly sure what the counterexample you were aiming at there is.

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## [SOLVED] Inverse Image of a Compact Set -- Bounded?

Oops - you're right of course. I had misinterpreted the question!

But anyway, there are lots of counterexamples - take for instance any constant function from R to R (in the standard metric).
 Recognitions: Gold Member Science Advisor Staff Emeritus What happens if you use a constant function? (specific example deleted- surely you can think of your own!)
 Ah, this is good stuff. Thanks folks. So I can choose some $$f:\mathbb{R} \rightarrow \mathbb{R}$$ such that $$\{f(x)=0 \mid x\epsilon\mathbb{R}\}$$, for instance. Then $$f$$ is compact easily since an open cover for it is $$\mathbb{R}$$ and a finite subcover is obvious. Then any $$f^{-1}$$ would send this $$\{0\}$$ back to $$\mathbb{R}$$, which is unbounded. Thanks again! This is solved.
 Recognitions: Homework Help Science Advisor That is odd notation. f is a function, not a subset of R. It is not compact. The image of f is a compact subset of R. However, the definition of compact is that *any* open cover has a finite subcover. You only wrote down one cover, that contained one set, not an arbitrary open cover. But a one point set is trivially compact, anyway, just not for the reasons you set out.

 Quote by matt grime That is odd notation. f is a function, not a subset of R. It is not compact. The image of f is a compact subset of R. However, the definition of compact is that *any* open cover has a finite subcover. You only wrote down one cover, that contained one set, not an arbitrary open cover. But a one point set is trivially compact, anyway, just not for the reasons you set out.
Ah, yes, all true. However, a union of all open covers should be $$\mathbb{R}$$. And $$f$$ was a typo, should be "the image of $$f^{-1}$$".

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 Quote by varygoode However, a union of all open covers should be $$\mathbb{R}$$.
What does that have to do with anything?

To show that {0} is compact you have to take an arbitrary open cover of {0} and show that it admits a finite subcover. But this is easy: at least one of the sets in any open cover will contain {0}, and any single such set will be a sufficient finite subcover.
 Recognitions: Gold Member Science Advisor Staff Emeritus Using f-1({0}) instead of f in your earlier response, that is simply R= f-1({0}) which is definitely NOT compact! In fact, you say that solves your problem because R is not bounded. Surely you know that every compact set, in a metric space, is bounded!
 HallsofIvy, it is only the {0} that needs to be compact, not $$f^{-1}(\{0\})$$. So it's all good.

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 Quote by varygoode HallsofIvy, it is only the {0} that needs to be compact, not $$f^{-1}(\{0\})$$. So it's all good.