Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with linear-orders
Search options not deleted
user 102946
19
votes
1
answer
1k
views
Suprema of directed sets
Let $(X, \le)$ be a partially ordered set. We call a subset $S \subseteq X$...
... a chain if each two elements in $S$ are comparable with respect to $\le$ (in other words, $S$ is linearly ordered wi …
- 12.5k
31
votes
Accepted
Characterizing $\mathbf{R}$ as an ordered group
The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following notatio …
- 12.5k