Suppose S is a bounded set and A is a subset of S. Prove A is bounded.

Question

terenzreignz · Answer

Y u no tag??
Anyway, I'll let you work this out yourself, I'll just nudge you in the right directions.
Trust me, it feels a lot better when you prove it yourself.

So, what are the facts?
S is bounded.  Then...?

terenzreignz · Answer

(I actually expounded on the boundedness of S in the first few lines of my proof for your previous question, you could actually just go ahead and copy-paste it here :P )

anonymous · Answer

I was about to tag. Promise. But otayyyy here i go... 
Since S is bounded, then there exist a and b such that for all s in S 
s ≤ b (its upper bound) 
and s ≥ a (its lower bound) 

Similarly, since T is bounded, then there exist c and d such that for all t in T 
t ≤ d (its upper bound) 
and t ≥ c (its lower bound) 

so far it must be right because it is your work

terenzreignz · Answer

Why did you drag T into all of this, there is no T in your question :P

terenzreignz · Answer

Scratch the T part, ok? :P

anonymous · Answer

i was JUST about to say that lol

anonymous · Answer

my bad. Ok. 
So since the statement that A=B means A is a subset of B, can I say S=A. And if we assumed S is bounded, then A is assumed to be bounded as well?....

terenzreignz · Answer

Whoa there.  To show that an entire set is, well, fits some characteristic (say, being bounded)
Choose an arbitrary element of that set, you better not name (specify) that element, and show that it does, indeed, fit the characteristic.

For instance, here's the frame of the proof.
Let a be an element of A.  Show that it is always less than or equal to some value, and that it is also greater than or equal to some other value.

anonymous · Answer

what value?

terenzreignz · Answer

Ok, I really hope you can understand every bit of this: Here we go:
S is bounded, then there exist a and b such that for all s in S 
s ≤ b (its upper bound) 
and 
s ≥ a (its lower bound) 

That said: Let A be a subset of S.  Let a be in A.
What is true about a?

anonymous · Answer

it is in S

terenzreignz · Answer

PRECISELY.  And being in S, what more can you say about it?

anonymous · Answer

its bounded...

terenzreignz · Answer

Let's backtrack a bit, I made the mistake of declaring a to be in A
when I already defined a to be the lower bound of S
gah

Let x be in A, instead.

anonymous · Answer

ok so x is in S. i got that.

terenzreignz · Answer

Ok, since x is in S, what properties hold for ALL things in S?

anonymous · Answer

they are bounded by a and b?

terenzreignz · Answer

that's right :)
So, you took x (an arbitrary element of A) and showed that 
a ≤ x ≤ b

Therefore, A is bounded.  That's done :)

anonymous · Answer

You are amazing. By far my favorite person on here. Thank you so much! You didn't just give me an answer; you explained it in a way I could understand. Thanks again!

terenzreignz · Answer

No problem :)