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What is the relation between compactness, connectedness and continuous real-valued functions on $\mathbb{R}^n$, n > 1?

For example, 1- what is the relation between a compact set and boundedness of every continuous real valued function on it?

2- If a set on $\mathbb{R}^n$ is compact, is it bounded? and if it is bounded is it compact?

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  • $\begingroup$ All this in the real line or something else?..your question lacks some context..you have to be more clear $\endgroup$ – Marios Gretsas Aug 20 '17 at 19:40
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    $\begingroup$ Do you mean theorems relating the various concepts? Well it's not hard to prove that the continuous image of compact spaces are compact and continuous image of connected spaces are connected. $\endgroup$ – John Griffin Aug 20 '17 at 19:41
  • $\begingroup$ On $\mathbb{R^n}$ @MariosGretsas $\endgroup$ – Emptymind Aug 20 '17 at 19:41
  • $\begingroup$ then see the comment of @JohnGriffin $\endgroup$ – Marios Gretsas Aug 20 '17 at 19:43
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    $\begingroup$ @Intuition I've added an answer addressing this question. Please edit your post to clarify what you're asking. $\endgroup$ – John Griffin Aug 20 '17 at 19:48
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Suppose $X$ is compact and $f:X\to\mathbb{R}^n$ is continuous. Then $f$ is bounded.

Proof: Since the continuous image of compact spaces are compact, we know that $f(X)$ is a compact subset of $\mathbb{R}^n$. Thus it is a bounded subset of $\mathbb{R}^n$ by the Heine-Borel theorem.

The Heine-Borel theorem mentioned above says:

A subset $E$ of $\mathbb{R}^n$ is compact iff $E$ is closed and bounded.

Thus compact subsets of $\mathbb{R}^n$ are bounded. However, the converse is not true. For instance, $(0,1)$ in $\mathbb{R}$ is bounded but not compact.

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  • $\begingroup$ Is it true that "If every continuous real valued function defined on a nonempty subset of $\mathbb{R^n}$", n>1, is bounded, then this non empty subset is compact? $\endgroup$ – Emptymind Aug 20 '17 at 23:53
  • $\begingroup$ @Intuition Yes. $\endgroup$ – John Griffin Aug 20 '17 at 23:58
  • $\begingroup$ why ? ..... is there a theorem that said this? $\endgroup$ – Emptymind Aug 22 '17 at 7:29
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    $\begingroup$ @Intuition accept whatever you want..accept the answer that you believe it helps you the most..Don't worry.. If you accept John's answer,which of course is a nice answer,there would be no problem by me. $\endgroup$ – Marios Gretsas Aug 22 '17 at 12:18
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    $\begingroup$ @Intuition See here: math.stackexchange.com/questions/668905/… $\endgroup$ – John Griffin Aug 22 '17 at 14:19
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A continuous image of a connected set is connected.

Also a topological space $X$ is connected iff there does not exist a surjective continuous function from $X$ to $\{0,1\}$(or any set with two elements in the reals for instance).

Now for compactness-continuity we have also these results:

If $f:X \rightarrow Y$ is a continuous bijection and $X$ ia compact topological space and $Y$ is a Hausdorf topological space then $f$ is a homeomorphism.

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Every continuous function on a compact metric space $X$ is uniformly continuous on $X$

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  • $\begingroup$ can I say that if a subset of $\mathbb{R^n}$, where n > 1, is compact then its connected? $\endgroup$ – Emptymind Aug 20 '17 at 22:49
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    $\begingroup$ No because the set $([0,1] \times [0,1]) \cup( [5,6] \times [5,6])$ is compact but not connected..Note that a finite union of compact sets is compact. $\endgroup$ – Marios Gretsas Aug 20 '17 at 22:54
  • $\begingroup$ I have a problem ..... your answer contains half of the answer of the question and @John Griffin contains the other half ..... so I do not know who to accept? $\endgroup$ – Emptymind Aug 22 '17 at 7:31

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