# Path connected compact set with given property

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.

• Did you mean $p: K \mapsto [0,1]$. I feel this is a trivialization of Tietze extension theorem – Yujie Zha Mar 15 '17 at 2:07
• @YujieZha no I don't think so. – IntegrateThis Mar 15 '17 at 2:09
• Ok, I'm still trying to understand your question. But then continuous function over a compact set would attain its max and min there. Thus $f(K)=[f(k_{min}),f(k_{max})]$ (see this link. I do not see the use of p here. @TheMathNoob – Yujie Zha Mar 15 '17 at 2:25
• @YujieZha good point, I am just the messenger of the question, will report back with more insights. – IntegrateThis Mar 15 '17 at 2:33

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.]

• Can you elaborate on what you mean by connected? Thanks. – IntegrateThis Mar 15 '17 at 2:18
• 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 2:19
• 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 2:22
• 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 – Yujie Zha Mar 15 '17 at 2:44
• @YujieZha posted my answer below, hopefully it is correct. – IntegrateThis Mar 15 '17 at 2:56

My answer in case anyone is interested. I do reference a theorem in my notes though.

• Looks good. You say that $f(K)$ is closed and bounded, so it might be good to spell out why the bounds are actually attained. (Really, this is the content of the extreme value theorem.) – Kenny Wong Mar 15 '17 at 3:01
• @KennyWong by Extreme Value them. – IntegrateThis Mar 15 '17 at 3:03
• That's exactly it! :-) – Kenny Wong Mar 15 '17 at 3:03
• Cool, looks good! (+1). One small suggestion: if you could edit it directly instead of putting it into picture, it'll be even better - so it will display better, and good for search. – Yujie Zha Mar 15 '17 at 3:30