Is the Inverse Image of a Compact Set Always Bounded?

In summary, the conversation discusses whether f^{-1}(K) is bounded when f is a continuous mapping from metric spaces X to Y and K \subset Y is compact. The conversation also mentions Theorem 4.8 Corollary (Rudin) which states that a mapping f is continuous if and only if f^{-1}(C) is closed in X for every closed set C in Y. The speaker suggests starting by describing K and f^{-1}(K) and considering why it might not be true before attempting to prove it. An example of f(x)=sin(x) is mentioned as a potential counterexample.
  • #1
bertram
2
0

Homework Statement



Let f be a continuous mapping from metric spaces X to Y. [tex]K \subset Y[/tex]is compact. Is [tex]f^{-1}[/tex](K) bounded?

Homework Equations


Theorem 4.8 Corollary (Rudin) A mapping f of a metric space X into Y is continuous iff [tex]f^{-1}[/tex](C) is closed in X for every closed set C in Y.

The Attempt at a Solution


So my idea was to show that [tex]f^{-1}[/tex](K) was continuous, but i can't really figure that out immediately.
I just tried next to describe K and [tex]f^{-1}[/tex](K) as best I could... We know that K is closed and compact (compact subsets of metric spaces are closed). This will imply that [tex]f^{-1}[/tex](K) is closed (Thm 4.8 corollary). So I have that K is closed and compact and that [tex]f^{-1}[/tex](K) is closed. I just don't know how to make the ends meet. Maybe I'm doing this wrong, or just missing something obvious.

Thanks in advance for any help.
 
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  • #2
Think about why it might not be true before you start trying to prove it. Suppose f:R->R and f(x)=sin(x)?
 

1. What is the inverse image?

The inverse image, also known as the preimage, is the set of all elements in the domain that map to a given element in the range of a function. In other words, it is the set of all inputs that produce a particular output.

2. How is the inverse image related to boundedness?

The inverse image can help determine if a function is bounded or not. If the inverse image is bounded, it means that for every output value, there is a finite number of input values that produce it. Conversely, if the inverse image is unbounded, there is an infinite number of input values that produce a particular output value.

3. What does it mean for the inverse image to be bounded?

If the inverse image is bounded, it means that for every output value in the range of the function, there is a finite number of input values that produce it. This also means that the function itself is bounded, as it has a finite range of output values.

4. How can we determine if the inverse image is bounded?

To determine if the inverse image is bounded, we can analyze the domain and range of the function. If the function has a finite range, then the inverse image will also be bounded. Additionally, we can graph the function and observe if there are any asymptotes or sharp turns, which may indicate an unbounded inverse image.

5. Why is the boundedness of the inverse image important?

The boundedness of the inverse image is important because it can help us understand the behavior of a function. If the inverse image is bounded, it means that the function has a finite range of outputs, which can have implications in fields such as economics, physics, and engineering. On the other hand, an unbounded inverse image can indicate that the function has infinite or undefined values, which may have different implications depending on the context of the function.

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