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Equip $\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\mathbb{R}$ is compact iff it's closed and bounded.

Is every closed bounded connected subset of $\mathbb{R}$ a closed interval $[a, b]$ (and conversely)?

Is every open bounded connected subset of $\mathbb{R}$ an open interval $(a, b)$ (and conversely)?

Is this somehow related to the fact that removing one point from $\mathbb{R}$ splits it into 2 disconnected pieces (how is this property called anyway)?

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  • $\begingroup$ All true if you agree to refer to a point as a closed interval. $\endgroup$ – Mikhail Katz Apr 24 '17 at 17:03
  • $\begingroup$ Do you consider $[b,b]= \{b\}$ to be a closed interval? $\endgroup$ – fleablood Apr 24 '17 at 17:04
  • $\begingroup$ No, I'm considering only non-trivial intervals. $\endgroup$ – étale-cohomology Apr 24 '17 at 17:04
  • $\begingroup$ beware also of the title, $[0,+\infty[$ is a closed interval but not bounded. $\endgroup$ – zwim Apr 24 '17 at 17:06
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    $\begingroup$ Do you want an informal or formal answer? It's intuitively obvious that only singletons and intervals are connected. And intuitively obvious that among finite intervals that only [a,b] are closed (and always closed) and (a,b) are open (and always open), and thus, yes, your statements are all true. But showing these formally via definitions is ... not hard, but tedious.... but good practice. $\endgroup$ – fleablood Apr 24 '17 at 17:30
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Clearly a closed and bounded interval is compact and connected.

Conversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \in I$, if $x < z<y$, then $z \in I$. All the sets with this property must have one of the forms $$\{a\},\,[a,b],\, ]a,b[, \,[a,b[, \,]a,b], \,]a,+\infty[,\, [a,+\infty[,\, ]-\infty,b[ \mbox{ or } ]-\infty,b].$$ Among these, only $\{a\}$ and $[a,b]$ are compact, hence the answer to your question is yes. Notice that $\{a\}$ is an interval, by definition.

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  • $\begingroup$ Thank you. What about the 2nd question, about open sets? $\endgroup$ – étale-cohomology Apr 24 '17 at 17:14
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    $\begingroup$ Also yes, with the same reasoning: use connectedness to narrow the list of "suspects" and then see that the only open bounded set in that list is $]a,b[$. $\endgroup$ – Ivo Terek Apr 24 '17 at 17:23
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Lemma 1: Only intervals and singletons and the empty set are connected and all intervals and singletons and the empty set are are connected.

Lemma Z: $K\subset \mathbb R$ is a interval if and only if for all $x,y \in K$ then for all $k; x < k < y; k \in K$.

Proof: Should be self-evident. If $K$ is an interval than $K = [(a,b)]$ (for sake of notation $a$ can be $-\infty$ and $b$ can be $\infty$). and $a \le x < y \le b$ and for all $k: x < k < y$ then $a < k < y$ so $k \in K$.

If there exists a $k$ so that $x < k < y$ with $k \not \in K$ and $x,y \in K$ then there is no $a,b$ (not even $\pm \infty$) so that $a \le x; b \ge y$ and for all $r \in \mathbb R$ $a < r < b$; $r \in K$. (as $a < k < b$ but $k \not \in K$). So $K$ would not be an interval.

Proof of Lemma 1: If $K$ = $\emptyset$ or $K = \{x\}$, some singleton then $K$ can not be partitioned into two partitions so $K$ is connected.

Let $E \subset \mathbb R$. And let $E$ be such that there exist $x,y\in E; x< y $ and there exists a $k \not \in \mathbb R$ so that $x < k < y$. Let $A = E \cap (\infty, k)$ and $B= E \cap (k, \infty)$ then $A, B$ are non-empty partitions of $E$ and $\overline A \subset (-\infty,k]$ is disjoint from $B \subset (k,\infty)$ and $\overline B \subset [k,\infty)$ is disjoint from $A \subset (-\infty, k)$ so $E$ is not connected.

So the only connected subsets of $E$ are intervals, singletons, and the empty set.

If $K$ is an interval and $K = A\cup B$ and $A,B$ non empty and $A \cap B = \emptyset$. Let $a \in A$ and $b \in B$ and wolog $a < b$. Let $K = \{x| x \ge a; \forall k; a\le k \le x: k \in A\}$ and let $L = \{y \in K| y > a; y \in B\}$. It's easy to prove $K$ is non-empty $(a \in K)$ and bounded above (by $b$) and $ L$ is non-empty ($b$ is in it) and bounded below (by $a$) and that $\sup L = \inf L$. As $K$ is an interval and $a \le j = \sup K = \inf L\le b$ then $j \in K$. And $j \in \overline A$ and $k \in \overline B$. So either $j \in \overline A \cap B$ or $j \in \overline B \cap A$. So $K$ is connected.

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  • $\begingroup$ My edit was solely to add a missing dollar-sign that disrupted some of the formatting. $\endgroup$ – DanielWainfleet Apr 25 '17 at 13:07

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