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UPSC 2013 Maths Optional Paper 2 Q2d — Step-by-Step Solution

14 marks · Section A

Real number system as ordered field with LUB property · Real Analysis · asked 2× in 14 yrs · Read the full method →

Question

Show that every open subset of $\mathbb R$ is a countable union of disjoint open intervals.

Technique

Connected-components construction via maximal open intervals; rationals provide the countability.

Solution

Let $U\subseteq\mathbb R$ be open. The proof proceeds in three steps: (1) construct the connected components of $U$ as a partition into maximal open intervals; (2) show distinct components are disjoint; (3) show there are countably many via the rationals.

Step 1 — Maximal open interval in $U$ through each point

Fix $x\in U$. Define

(If the set has no lower bound, $\alpha_x=-\infty$; similarly $\beta_x=\infty$ is allowed.)

Then $\alpha_x<x<\beta_x$ (because $U$ is open: a neighbourhood $(x-\epsilon,x+\epsilon)\subset U$ exists, so $\alpha_x\le x-\epsilon<x<x+\epsilon\le\beta_x$).

Set $I_x=(\alpha_x,\beta_x)$ — open interval (possibly unbounded).

Claim: $I_x\subset U$, and $I_x$ is the largest open interval in $U$ containing $x$.

Proof. For any $y\in I_x$, WLOG $y>x$. By definition of $\beta_x$ as a sup, for $\epsilon=\beta_x-y>0$ there exists $b$ with $[x,b)\subset U$ and $b>\beta_x-\epsilon=y$. So $y\in[x,b)\subset U$. Similarly for $y<x$. Hence $I_x\subset U$.

Conversely, if $J\subset U$ is any open interval containing $x$, then $J\subset I_x$ by definition (any open interval through $x$ contained in $U$ stays within the bounds $\alpha_x,\beta_x$).

Step 2 — Distinct components are disjoint

Claim: For $x,y\in U$, either $I_x=I_y$ or $I_x\cap I_y=\emptyset$.

Proof. Suppose $I_x\cap I_y\ne\emptyset$. Pick $z\in I_x\cap I_y$. Then $I_x\cup I_y$ is an open interval containing $z$ (the union of two overlapping open intervals is an open interval) and contained in $U$. So $I_x\cup I_y\subset I_z\subset I_x$ (by maximality of $I_x$ at $z$, since $I_x$ contains $z$). Hence $I_y\subset I_x$, and similarly $I_x\subset I_y$, so $I_x=I_y$.

Step 3 — Countability

The components $\{I_x:x\in U\}$ (collected with each distinct $I_x$ taken once) partition $U$ into pairwise disjoint open intervals.

For each component $I$, $I$ is a non-empty open interval, so $I\cap\mathbb Q\ne\emptyset$ (rationals are dense in $\mathbb R$). Pick $q_I\in I\cap\mathbb Q$.

Different components are disjoint, so different components contribute different rationals: $I\ne I'\Rightarrow q_I\ne q_{I'}$.

The map $\{I\}\to\mathbb Q$, $I\mapsto q_I$, is injective. Since $\mathbb Q$ is countable, the set of components is countable.

Conclusion

Answer