Doubt 1: August 18, 2026
OKAY! Let’s attempt this.
The Theorem says:
Theorem
Let \(X \subseteq \mathbb{R}\) be non-empty and bounded. Then, \(M_X^* = \sup X\) iff:
- \(M_X^*\) is an upper bound of \(X\).
- \(\forall \varepsilon > 0, \exists x' \in X\), \(M_X^* - \varepsilon < x' \le M_X^*\).
Before beginning with the proof, let’s get back to the definitions of Supremum, upper bound, and bounded set.
- Bounded Set: A set \(X\) is bounded if there is a real \(M > 0\) such that \(\forall x \in X\), \(\vert{}x\vert{} < M\).
- Upper Bound: A set \(X\) is said to be bounded above, if there exists a real \(\beta\) such that \(x \le \beta\) for every \(x \in X\), and we call \(\beta\) an upper bound of \(X\).
- Least Upper Bound (Supremum): Suppose \(E\) is bounded above, and there exists an \(\alpha\) with the following properties:
- \(\alpha\) is an upper bound of \(X\).
- If \(\gamma < \alpha\), then \(\gamma\) is not an upper bound of \(X\).
Then we call \(\alpha = \sup(X)\).
Understanding “If and Only If” (iff) Statements
Now, that we recalled the defs, let’s look at how do we prove iff (if and only if) statements.
Iff statements are written as:
\[A \iff B\]This statement is equivalent to:
\[(A \implies B) \land (B \implies A)\]We can check the truth table:
| \(A\) | \(B\) | \(A \iff B\) | \(A \implies B\) | \(B \implies A\) | \((A \implies B) \land (B \implies A)\) |
|---|---|---|---|---|---|
| T | T | T | T | T | T |
| T | F | F | F | T | F |
| F | T | F | T | F | F |
| F | F | T | T | T | T |
So we need to prove both \((A \implies B) \land (B \implies A)\) to be true statements
Applying the Structure to the Theorem
Let’s revisit the theorem again,
Let \(X \subseteq \mathbb{R}\) be non-empty & bounded (Assumption). Then,
\[\underbrace{M_X^* = \sup X}_{A}\]iff
\[\underbrace{\text{(i) } M_X^* \text{ is an upper bound of } X}_{B}\] \[\underbrace{\text{(ii) } \forall \varepsilon > 0, \exists x' \in X, M_X^* - \varepsilon < x' \le M_X^*}_{B}\]Proof
We must prove both directions of the “if and only if” statement: \((A \implies B)\) and \((B \implies A)\).
Part 1: Proving \(A \implies B\) (“Only If”)
Assumption (\(A\)): Let \(M_X^* = \sup X\). To Show (\(B\)):
- \(M_X^*\) is an upper bound of \(X\).
- \(\forall \varepsilon > 0, \exists x' \in X\) such that \(M_X^* - \varepsilon < x' \le M_X^*\).
- Proof of Condition 1: By definition, the supremum of a set is its least upper bound, which inherently means \(M_X^*\) is an upper bound of \(X\). Thus, condition (1) holds automatically.
- Proof of Condition 2: Let \(\varepsilon > 0\) be given. Because \(M_X^*\) is the least upper bound (\(\sup X\)), any number strictly less than \(M_X^*\) cannot be an upper bound of \(X\). Since \(\varepsilon > 0\), we have:
Therefore, \(\gamma\) is not an upper bound of \(X\). By the definition of an upper bound, for \(\gamma\) to fail to be an upper bound, there must exist some element \(x' \in X\) that exceeds it:
\[M_X^* - \varepsilon < x'\]Furthermore, since \(M_X^*\) is an upper bound of \(X\), every element in \(X\) is less than or equal to \(M_X^*\), meaning \(x' \le M_X^*\). Combining these gives:
\[M_X^* - \varepsilon < x' \le M_X^*\]This completes the proof of Part 1.
Part 2: Proving \(B \implies A\) (“If”)
Assumption (\(B\)):
- \(M_X^*\) is an upper bound of \(X\).
- \(\forall \varepsilon > 0, \exists x' \in X\), \(M_X^* - \varepsilon < x' \le M_X^*\).
To Show (\(A\)): \(M_X^* = \sup X\).
The strategy is to use contradiction and assume that \(M_X^* \neq \sup{X}\). If \(M_X^* \neq \sup{X}\), then there must be some other number less than \(M_X^*\) which would be the supremum. We can assume that number to be \(\gamma\). Using the given assumptions in (B), we would finally show that \(\exists x' \in X\) such that \(x'> \gamma\) and so \(\gamma\) cannot be supremum.
- Proof: We already know from assumption (1) that \(M_X^*\) is an upper bound of \(X\). To show that \(M_X^*\) is the least upper bound (\(\sup X\)), we must show that no number strictly less than \(M_X^*\) can be an upper bound. We proceed by contradiction. Assume that \(M_X^*\) is not the supremum. Then there exists some upper bound \(\gamma\) of \(X\) such that:
Let us define \(\varepsilon = M_X^* - \gamma\). Since \(\gamma < M_X^*\), it follows that \(\varepsilon > 0\). By our assumption (2), because \(\varepsilon > 0\), there exists an element \(x' \in X\) such that:
\[M_X^* - \varepsilon < x'\]Substituting our definition of \(\varepsilon\) = (\(M_X^* - \gamma\)) back into the inequality gives:
\[M_X^* - (M_X^* - \gamma) < x' \implies \gamma < x'\]The Contradiction: We have found an element \(x' \in X\) that is strictly greater than \(\gamma\) (\(\gamma < x'\)). This directly contradicts the assumption that \(\gamma\) is an upper bound of \(X\). Therefore, no upper bound can be strictly less than \(M_X^*\), proving that \(M_X^*\) is indeed the least upper bound. Thus, \(M_X^* = \sup X\). We proved \(B \implies A\).