Veeky Forums, is this true or false? I think it is true

Veeky Forums, is this true or false? I think it is true.

let A be R
B be Q
and C be Q aswell
false

Fuck I fucked that shit up. This is the correct one, that B should have been a C

Still false.

C = {1}
A = {1,2}
B = {1,3}

Yeah but A is still contained in B though? Or does the entire set need to be contained in it? I'm just trying to confirm because I just got this test back and I got a 56

>Thinking {1,2} is contained in {1,3}
I can see why you got 56

where is the 2 in B

fuck off already

Yeah, I fucked up in my thinking when I was taking it. I deserve the 56 but this is some shit. That's probably the lowest grade I've ever gotten on a test...

One more question.

It states that every finite set of real numbers contains its supremum.

Using interval notation the set [1) contains only one element yet it doesn't contain its supremum.

Would it be worthwhile to contact the publisher to tell him he's made a mistake?

>[1) contains only one element
No it doesn't.
Let a be in [1)
1

No.

Consider

[eqn] A = \{ 1,2,3 \}, \;\; B = \{ 1,2,4,5 \}, \;\; C = \{ 1,2,4,6 \}. [/eqn]

We have immediately that

[eqn] A \cap B = \; \{ 1,2 \}, \;\; B \cap C = \; \{ 1,2,4 \}, \;\; A \cap B \subseteq B \cap C. [/eqn]

But A is not a subset of B, because the element 3 is not a member of B. Therefore the statement proposed in the OP, as it pertains to all sets in general, is false. The conclusion does not necessarily follow from the premise, as has just been shown by the above counterexample.

Notice that in constructing the above counterexample, I chose A,B and C to illustrate a situation where the two "intersected" sets were not precisely equal to one another. Some authors use the subset symbol used in the OP to connote the situation where one set is either completely contained in the second (with elements left over in the second), or otherwise in the case where the two sets are precisely equal. Some authors reserve the other symbol, without the horizontal bar on the bottom, like the bottom half of ">=", "=" etc, specifically for the former situation, where the two sets are /not precisely equal/. These choices of usage are not uniform.

Exercises: consider the above, and modify the counterexample to give a situatoin where the two intersected sets are precisely equal. Does this change anything? Why or why not? Also, use the above type of example to solve the corrected problem, if possible (feel free to construct your own sets to do this).

Top tier autism mate, especially since you took 5x the time to do what already did. And the fact you didn't read any post at all, including the one that clarified an error in the pic in the OP.

Are you a teacher, by chance?

It does contain one element you fucking retard. That's what the "[" denotes, you're probably an engineer, leave real maths to mathematicians sweety.

>question gets clearly answered in 3 minutes and 10 words
>better spend an hour writing out paragraphs of rambling explanation
>better assign some fucking homework along with my Veeky Forums post too

On the contrary, it's you who can't read.

If you read the post (I know that reading more than a few dozen words at a stretch is too much to ask), then you'll clearly see that I was aware that the OP "corrected" the problem, thus presenting a distinct problem, a few posts later. But you can't admit that you're wrong at this point, because you're dug into a position in an Internet Argument that you have to win at all costs, or something.

Let [math]n\in[1),\ n\geq1,\ n

>I was aware that the OP "corrected" the problem, thus presenting a distinct problem, a few posts later.
Apparently that's where you stopped because the corrected problem was also answered an hour ago in the very next post down

Yeah I see I'm wrong now about that. I don't see why you even bothered to write all that though, as is pretty much the same as yours, and OP already understood what he did wrong ( and , and the fact he added an additional question ).

I thought you might reply in one of a few different ways:

either not actually re-read my post and blow it off, emphasizing that "the post is still retarded" or something along these lines,

or else: actually re-read my post, privately think "fuck he got me", and then not concede the point, instead just doubling down in the same way as above.

You've instead chosen a third way (which I honestly hadn't expected), which is interesting in the sense that it's actually argumentatively the weakest of the three, so far: you did privately realize "fuck he got me", but you have to continue to imply that I haven't been reading the thread, because that's how you started out, and so that's one area for you to double down on. Except you just tacitly admitted that I /did/ read the thread, up to a point, which just undermines your new, weaker argument.

The argument is made weaker still by the simple fact that both the OP goof and the correction are both things that can be discussed, regardless of the OP's intent.

AH, a polite admission! gj, acknowledged. Another thing I didn't expect.

In return, I have an admission of my own (I must be careful here): I took to be the same person as , which might be true, but is not necessarily so, in light of this latter "admission" post. Whoever the authors, the thrust of the post that I snarked back at is still wrong, for basically the reasons that I gave.

>That's probably the lowest grade I've ever gotten on a test
>literally university math 101 week 1

you talk like you own an entire rack of fedoras

>inb4 he posts a 10 page essay trying to analyze you

No ik, that's why I feel like shit. This is in a proofs class and for some reason my intuition isn't as strong.

Are you serious?
[1) is the empty set.

>Would it be worthwhile to contact the publisher to tell him he's made a mistake?
If you seriously think you noticed something which the publisher didn't notice, you are wrong.


kill yourself.

What if a and b is empty

This is clearly false. Here is a counterexample.

A = {1}
B = { }
C = {2}

Then clearly

[math]
A \cap B \subseteq B \cap C
[/math]

However, A is not a subset of B.

As a friendly advise, you shouldn't be really taking these courses.

It's true, because empty set is a subset of every other set

/thread