Anki: separating a large deck with many overdue cards into two piles

Just due:

deck:"deckname" is:due prop:due>-7

Overdue:

deck:"deckname" is:due prop:due<=-7

Of course one must replace deckname with one’s own deck.

I’ve also decided to split my collection under two accounts, one for regular decks (which is synced with Ankiweb) and one for media-intensive decks (which remains local). This solved the problem I was having with syncing media on my Android phone, but I am afraid I will be less willing to review my unsynced decks.

EDIT: I’ve realized that the other problem I was having with syncing was my unstable WiFi connection (my phone seems to disconnect from the WiFi very frequently, for some reason). Anki doesn’t seem to be able to deal with my specific connection problems, and thus when I tried to sync when it was disconnected (though nominally connected), it would throw out errors. My crude solution is to start Ankidroid, then turn off WiFi, then restart WiFi, so that the WiFi is definitely on, then finally sync.

EDIT2: Ankidroid still seems to be having problems, despite the trick above…

Two definitions for the least upper bound

The definition for the least upper bound seems to be given in two nearly identical ways, and I wanted to show that these definitions are, in fact, equivalent, so here is a little note.

Consider the following three statements regarding some set of real numbers , and real numbers .

1. for all .
2. Let for all ; then .
3. Let ; then there is some such that .

My impression is that most books use (1) and (2) above to define as the least upper bound of . (To me, at least, this seems most intuitive, since we actually “see” that out of all the upper bounds of , is indeed the least.) However, some books e.g. Ross in Elementary Analysis (p 21; it’s actually not part of the real definition, but (1) and (3) are the only “formal” parts given) seem to use (1) and (3) instead.

Our task, then, is to show that

• (1) and (2) together imply (3), and that
• (1) and (3) together imply (2).

Proof: To show the first implication, suppose (1) and (2) are true. We want to show (3). Let . We need to find such that . Suppose for the sake of contradiction that such does not exist, i.e., for all we have . Then by (2) we have . But , which is a contradiction.

Now suppose (1) and (3) are true; we would like to show (2). Let for all . We now need . Suppose for the sake of contradiction that , i.e., . Then there is, by (3), some such that . This contradicts the above, where was an upper bound for .