Question

Let \( P \) and \( Q \) be two non-empty disjoint subsets of \( \mathbb{R} \). Which of the following is (are) FALSE?

• A. If \( P \) and \( Q \) are compact, then \( P \cup Q \) is also compact
• B. If \( P \) and \( Q \) are not connected, then \( P \cup Q \) is also not connected
• C. If \( P \cup Q \) and \(P\) are closed, then \( Q \) is closed
• D. If \( P \cup Q \) and \(P\) are open, then \( Q \) is open

Hint:

When working with closed sets, remember that the closure of a union does not necessarily imply the closure of the individual sets.

Answer 1

The Correct Option is B, C, D

Given: $P$ and $Q$ are two non-empty disjoint subsets of $\mathbb{R}$.

Option (A): If $P$ and $Q$ are compact, then $P \cup Q$ is also compact

In $\mathbb{R}$, a set is compact if and only if it is closed and bounded.

If $P$ and $Q$ are both compact:

Both are closed and bounded

The union of two closed sets is closed

The union of two bounded sets is bounded

Therefore, $P \cup Q$ is closed and bounded, hence compact.

Option (A) is TRUE 

Option (B): If $P$ and $Q$ are not connected, then $P \cup Q$ is also not connected

Counterexample:

Let $P = [0,1] \cup [2,3]$ (disconnected) and $Q = (1,2)$ (connected).

Then $P \cup Q = [0,3]$, which is connected!

Actually, the statement says "If $P$ and $Q$ are not connected" - this means both are not connected.

Let $P = [0,1] \cup [3,4]$ and $Q = (1,3)$.

Then $P \cup Q = [0,1] \cup (1,3) \cup [3,4] = [0,4] \setminus {1,3} \cup {1,3}... $, this gives $[0,4]$ which is connected.

Actually, even if both $P$ and $Q$ are disconnected, their union can be connected if $Q$ "fills the gaps" in $P$.

Option (B) is FALSE 

Option (C): If $P \cup Q$ and $P$ are closed, then $Q$ is closed

Since $P$ and $Q$ are disjoint: $P \cup Q = P \sqcup Q$

Given: $P \cup Q$ is closed and $P$ is closed.

Since $P$ is closed and $P \subseteq P \cup Q$, we have: $$Q = (P \cup Q) \setminus P$$

The set $(P \cup Q) \setminus P$ is the set difference of a closed set and a closed set.

In general, the difference of two closed sets need not be closed.

Counterexample:

Let $P = {0}$ (closed) and $Q = (0, 1)$ (open, not closed).

Then $P \cup Q = {0} \cup (0,1) = [0,1)$.

But $[0,1)$ is NOT closed in $\mathbb{R}$! So this doesn't work as a counterexample.

Let me try: $P = (-\infty, 0]$ and $Q = (0, 1)$.

Then $P \cup Q = (-\infty, 1)$, which is not closed. So this doesn't satisfy the hypothesis.

Actually, if $P \cup Q$ is closed and $P$ is closed, then: $$Q = (P \cup Q) \cap P^c$$

Since $P$ is closed, $P^c$ is open. The intersection of a closed set and an open set can be either open, closed, or neither.

Counterexample:

Let $P = [0, 1]$ and $Q = (1, 2]$.

$P$ is closed ✓

$P \cup Q = [0, 2]$ is closed ✓

But $Q = (1, 2]$ is NOT closed (it doesn't contain its limit point 1, but 1 is in $P$, so it's in the disjoint set)

Wait, $Q = (1,2]$ is actually not closed because... actually, we need to check if $\overline{Q} = Q$.

$\overline{Q} = \overline{(1,2]} = [1,2]$, so $Q \neq \overline{Q}$, meaning $Q$ is not closed.

Option (C) is FALSE 

Option (D): If $P \cup Q$ and $P$ are open, then $Q$ is open

Given: $P \cup Q$ is open and $P$ is open.

Since $P$ and $Q$ are disjoint: $$Q = (P \cup Q) \setminus P = (P \cup Q) \cap P^c$$

Since $P$ is open, $P^c$ is closed. The intersection of an open set with a closed set can be open, closed, or neither.

Counterexample:

Let $P = (0, 1)$ and $Q = [1, 2)$.

$P$ is open ✓

$P \cup Q = (0, 2)$ is open ✓

But $Q = [1, 2)$ is NOT open

Option (D) is FALSE 

Answer: (B), (C), and (D)