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Problem: Suppose $K$ is a compact subset of $\Bbb R^n$, and that for all $k_1, k_2 \in K$, there exists a continuous function $p:[0,1] \rightarrow K$ such that $p(0) = k_1 $ and $p_1 = k_2$. Then let $f: K \rightarrow \Bbb R$ be continous on $K$. Prove that there exists $k_{min}, k_{max}$ such that $f(K) = [f(k_{min}), f(k_{max})]$.

Thoughts: First since $K$ is compact it is closed and bounded. Next I know that I have to show two inclusions, namely that $f(K) \subset [f(k_{min}), f(k_{max})]$ and the reverse inclusion.

I also know that the composition $f(p(x))$ is continuous for $x \in [0,1]$, and I wish I could post more but I am very stuck. Hints much appreciated (emphasis that I want a hint and not a solution please). I am also aware of basic theorems like Extreme Value, Intermediate value etc.

IntegrateThis
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2 Answers2

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The first condition you're given tells you that $K$ is path connected. Therefore $K$ is connected.

You are also told that $K$ is compact.

The image of a connected (resp. compact) space under a continuous map is connected (resp. compact). So $f(K) \subset \mathbb R$ is both connected and compact.

Now ask yourself: what are the only connected, compact subsets of $\mathbb R$?

[Alternatively, since you mentioned the intermediate value theorem and the extreme value theorem in your post, observe that both of these theorems are applicable to your map $f$, since $K$ is both connected and compact.]

Kenny Wong
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  • Can you elaborate on what you mean by connected? Thanks. – IntegrateThis Mar 15 '17 at 02:18
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    Think of a connected set as a set on which the intermediate value theorem is true. More geometrically, it's a set with "no gaps". So $[0,1]$ is connected and $[0,1) \cup (2, 3]$ is disconnected. The rigorous definition is that a space $S$ is connected iff it is impossible to write $S$ as a disjoint union of two proper open subsets. – Kenny Wong Mar 15 '17 at 02:19
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    If you're not familiar with connectedness, then perhaps you could apply the intermediate value theorem to $f \circ p$ to show that the $f(K)$ has "no gaps". – Kenny Wong Mar 15 '17 at 02:22
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    Thanks @KennyWong (+1) for your answer. Yea, the first condition is used to make sure K is connected, otherwise, $f(K)$ could not be an entire interval. $f$ is continuous, but that's in terms of $K$, so that does not guarantee $f(K)$ is an entire interval in $\mathbb R$, but the continuous mapping helps to make sure $f(K)$ is connected. And in terms of subset of $\mathbb R$ has to be an interval, one reference could be Prop 20.47 of Real Analysis by Richard – Jay Zha Mar 15 '17 at 02:44
  • @YujieZha posted my answer below, hopefully it is correct. – IntegrateThis Mar 15 '17 at 02:56
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My answer in case anyone is interested. I do reference a theorem in my notes though.enter image description here

IntegrateThis
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