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In this question What is the difference between minimum and infimum?, the answer of Thomas contains a line that

It is a fact that every set of real numbers has an infimum.

If $S=\text{{$-\infty,\ldots,0$}}$, does there exist infimum?

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user 31466
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  • Your definition of $S$ is unclear. – Git Gud Apr 25 '14 at 10:27
  • $-\infty$ is not a real number, but the statement should read: every $\textbf{bounded}$ set of real numbers has an infimum. – William Apr 25 '14 at 10:29
  • @GitGud By $S=\text{{$-\infty,\ldots,0$}}$, i wanted to mean that $S$ is a set such that it contains the elements from $-\infty,\ldots,0$. – user 31466 Apr 25 '14 at 10:30
  • I red the answer of Thomas and I'm not convinced. If you don't extend your order to $-\infty$ and lie in $\overline{\mathbb{R}}$, then $\mathbb{R}$ itself doesn't have an infimum. – yago Apr 25 '14 at 10:32
  • @Leaf That's usually denoted by interval notation: $(-\infty , 0]$ or $]-\infty , 0[$. – Git Gud Apr 25 '14 at 10:35

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Whether every set of real numbers has an infimum or not depends on the underlying partial order.

If the poset is $\left( \mathbb R,\leq\right)$ where $\leq$ is the usual order, then the answer is no. As a counterexample to the statement take the set $\{x\in \mathbb R\colon x\leq 0\}$, (note how the symbol $\infty$ doesn't come up). For it to be true you need to require that the set is bounded below.

If the poset is $\left( \mathbb R\cup \{\pm \infty\},\preceq\right)$ where $\preceq$ is an extension of $\leq$ given by $\forall x\in \mathbb R(-\infty\prec x\prec \infty)$, then the answer is yes, every subset of $\mathbb R$ has an infimum. This doesn't mean the infimum is a real number, though.

Git Gud
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The "number" $-\infty$ is not part of the standard real numbers. Normally it is added with $\infty$ as a convenience in mathematical analysis. Together with the real numbers they form the "extended real numbers" that is usually noted as $\overline{\mathbb{R}}$.

The element $-\infty$ satisfies that for any standard real number $x$, $-\infty < x$, so in your case $\inf S = -\infty$ and is indeed a minimum, because $-\infty \in S$, although not a very informative one ;).

Bunder
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  • If $S=]-\infty,0]$, then is that inf $S=-\infty$ indeed a minimum ? – user 31466 Apr 25 '14 at 10:40
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    @Leaf: $-\infty \not\in ]-\infty,0]$. By definition the minimum of a set belongs to that set, so $-\infty$ cannot be the minimum. – Najib Idrissi Apr 25 '14 at 10:52
  • @NajibIdrissi did Bunder consider $S$ as $[-\infty,0]$ or $[-\infty,0[$ . So he said "in your case $\inf S = -\infty$ and is indeed a minimum"? – user 31466 Apr 25 '14 at 10:58
  • Just a little typesetting advice: When you put a bar on large symbols like $\mathbb R$, the usual \bar is often too small, like $\bar{\mathbb R}$. You can use \overline to produce $\overline{\mathbb R}$. (I didn't see that bar when I first read your answer, it was very close to the "y" of "analysis" and tiny, so I missed it.) – Christoph Apr 25 '14 at 11:45
  • @Leaf: I just considered the fact that $-\infty \in S$. From your question I cannot deduce if $S = [-\infty, 0]$ or $S = [-\infty, 0[$ or anything else apart from the fact that $-\infty \in S$ and $0 \in S$. – Bunder Apr 25 '14 at 12:45
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It seems you caught me in stating an inaccuracy in my answer. Thanks for that.

So to clarify: As @GitGud says in his answer, I should have said that every non empty set that is bounded below has an infimum. We remember that $\infty$ and $-\infty$ are not real numbers.

If $S$ is unbounded below, then we can say that the infimum of $S$ is $-\infty$. Likewise, if the set $S$ is empty, then it is common to say that the infimum is $\infty$.

Edit: I wanted to try an answer the question in the comments below. If you have a set $T$ with a partial order, then we can talk about, for example, the infimum of a subset $S$. This infimum is the greatest lower bound for the set $S$. The point is that the infimum is an element in $T$. Now how does this work for $\mathbb{R}$? Usually we would consider $\mathbb{R}$ as the partially ordered set and we consider subsets of $\mathbb{R}$. That means that any infimum of any set of real numbers would be an element of $\mathbb{R}$, i.e. a real number. $\infty$ and $-\infty$ are not real numbers. And so, for example, the sets $\{x\in \mathbb{R}\mid x\leq 0\}$ or $\{\dots ,-3, -2, -1, 0\}$ do not have infimums. There is no element of $\mathbb{R}$ that is less than or equal to every element of either of those sets.

If your question you ask about $\{-\infty, \dots, -1, 0\}$. This set is not a set of real numbers though since $-\infty$ is not a real number.

But if you, as GitGud suggests consider the set as a subset of the partially ordered set $\mathbb{R}\cup\{-\infty, \infty\}$, then indeed you would have infimums.

In all it becomes a question of whether you consider your given set a subset of real numbers of as a subset of $\mathbb{R}\cup\{-\infty, \infty\}$.

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Thomas
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  • The @ thing doesn't work in answers. – Git Gud Apr 25 '14 at 11:44
  • I apologize. I have not understood yet. As far , i did know $\mathbb R$ is not bounded below and bounded above . So $\mathbb R$ have neither infimum nor supremum. Then if $S$ is unbounded below, then how does it have inf $S=\infty$? – user 31466 Apr 25 '14 at 11:51
  • @Leaf: Don't apologize. I should apologize for not be precise in my earlier answer. I have tried to answer your question above. – Thomas Apr 25 '14 at 12:44