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Why is the empty set bounded below and bounded above? If it has no elements, how can you say that an upper or lower bound exists?

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    $\begingroup$ Every number $b$ is both, an upper and a lower bound. For there is no $x \in \varnothing$ such that $x < b$ or $x > b$. $\endgroup$ – Daniel Fischer Oct 19 '13 at 16:25
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    $\begingroup$ Note that $17$ is an upper bound. For it is true that every element of the empty set is $\lt 17$. Can you name one that isn't? $\endgroup$ – André Nicolas Oct 19 '13 at 16:25
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    $\begingroup$ This is an instance of something being vacuously true. $\endgroup$ – Francis Adams Oct 19 '13 at 16:25
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    $\begingroup$ Definition: a subset $S$ of $\mathbb{R}$ is blue if there exists $s\in S$ such that $17s+\sqrt{3}$ is rational. Proposition: the empty set is not blue. $\endgroup$ – Julien Oct 19 '13 at 16:34
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Recall that implication has the property that when the assumption is false, the implication is true. In other words, if $P$ is false, then $P\implies Q$ is true.

Let $S$ be a set of real numbers. Then $M$ is an upper bound for $S$ if the following implication holds, $$s\in S\implies s \leq M.$$ Now let us examine the case for the empty set, $\emptyset$.

Proposition Let $M$ be any real number. Then $M$ is an upper bound for the empty set (of real numbers).

Proof:

Since the statement, $s\in\emptyset$ is false, the implication, $$s\in\emptyset\implies s \leq M$$ is true.

Note: An almost identical proof works for the lower bound case. A nice little slogan to remember here is

All things are true about the MEMBERS of the empty set.

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Argue by contradiction. Suppose $\emptyset$ is unbounded. Then for every $M > 0$ there is a point $x \in \emptyset$ such that $|x| > M$. But this contradicts that the empty set has no elements.

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Alternatively, a set $S$ is bounded if there exist numbers $a$ and $b$ such that: $$S\subseteq[a,b]$$ Now, $\varnothing$ is a subset of every set, so…

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A set in a metric space is bounded if, and only if, there exists a ball (of finite radius) containing it. Then trivially the empty set is contained in a ball (actually in every ball).

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In mathematical analysis and related areas of mathematics, a set is called bounded, if it is, in a certain sense, of finite size. Conversely, a set which is not bounded is called unbounded. - Wikipedia definition.

Now there is only one "nothing" just like there is only one 1, or only one 2, or say only one zero. There may be 2 cats or 2 dogs or 2 giraffes but 2 is 2. there is only one 2 along the entire number-line. similarly there are only one "nothing".

so the empty set, the content (nothing) is finite in number, which is 1. so empty set should be finite.

please correct me i may make mistake im not math people.

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    $\begingroup$ The quote you provide says that a set is bounded "if it is, in a certain sense, of finite size." This does not mean that the set is of finite cardinality. This is not the "certain sense" alluded to in the quote. $\endgroup$ – Xander Henderson Sep 1 '19 at 18:55
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We can say that the empty set is bounded if its not in R; that is if the empty set is the complement of R then we can say is bounded, because it lies in the extension real numbers.But if the empty set is a subset of R then it may be bounded or unbounded. If we assume it finite then its bounded set but if we look at its supremum and its infimum we see that they are not real numbers so the set is unbounded. I see that the empty set is both bounded and unbounded.

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  • $\begingroup$ This is a really bad way of looking at it. The supremum is $-\infty$; this means that any extended real number is an upper bound. Similarly the infimum is $+\infty$, which means that any extended real number is a lower bound. No other subset of $\mathbb{R}$ has either of these properties. In this sense the empty set is "more bounded" than any other set. $\endgroup$ – Ian Dec 9 '14 at 21:43

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