- #1

AGNuke

Gold Member

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The Question -

Let ##\mathbb{A}## be a non-empty subset of ##\mathbb{R}## and ##\alpha \in \mathbb{R}##. Show that ##\alpha = sup\mathbb{A}## if and only if ##\alpha - \frac{1}{n}## is not an upper bound of ##\mathbb{A}## but ##\alpha + \frac{1}{n}## is an upper bound of ##\mathbb{A}\;\forall \;n \in \mathbb{N}##.

My solution -

I tried to approach this completely obvious question (as per to me) with two contradictions - one for each case.

Let ##\alpha## be an upper bound but ##\alpha + \frac{1}{n}## isn't upper bound. Then, ##\exists \; a\in \mathbb{A}## such that

$$\alpha + \frac{1}{n} \leq a \Rightarrow a - \alpha \geq \frac{1}{n}\; \forall \;n \in \mathbb{N}$$

but from the first condition,

$$\alpha \geq a \;\forall\;a\in\mathbb{A} \Rightarrow \Leftarrow $$

Hence, our assumption is flawed. Hence ##\alpha - \frac{1}{n}## must be upper bound if ##\alpha## is to be the supremum of ##\mathbb{A}##.

Similarly for the second condition (by contradiction)

Help required -

The thing is, I simply can't get to solve such "simple" and "obvious" question in such methods without toiling for hours. If possible, can someone refer me to course materials where these "obvious" questions are solved by such "not-so-obvious" methods? I am really troubled and unable to understand things like these in class, furthermore, my exams are drawing near.

Let ##\mathbb{A}## be a non-empty subset of ##\mathbb{R}## and ##\alpha \in \mathbb{R}##. Show that ##\alpha = sup\mathbb{A}## if and only if ##\alpha - \frac{1}{n}## is not an upper bound of ##\mathbb{A}## but ##\alpha + \frac{1}{n}## is an upper bound of ##\mathbb{A}\;\forall \;n \in \mathbb{N}##.

My solution -

I tried to approach this completely obvious question (as per to me) with two contradictions - one for each case.

Let ##\alpha## be an upper bound but ##\alpha + \frac{1}{n}## isn't upper bound. Then, ##\exists \; a\in \mathbb{A}## such that

$$\alpha + \frac{1}{n} \leq a \Rightarrow a - \alpha \geq \frac{1}{n}\; \forall \;n \in \mathbb{N}$$

but from the first condition,

$$\alpha \geq a \;\forall\;a\in\mathbb{A} \Rightarrow \Leftarrow $$

Hence, our assumption is flawed. Hence ##\alpha - \frac{1}{n}## must be upper bound if ##\alpha## is to be the supremum of ##\mathbb{A}##.

Similarly for the second condition (by contradiction)

Help required -

The thing is, I simply can't get to solve such "simple" and "obvious" question in such methods without toiling for hours. If possible, can someone refer me to course materials where these "obvious" questions are solved by such "not-so-obvious" methods? I am really troubled and unable to understand things like these in class, furthermore, my exams are drawing near.

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