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inf (s) and sup (s)

S = {2x^2 : x ∈ R and x^3 > −8}I need help with determining the inf (s) and the sup (s) and whether theyre in the set or not. Thanks in advance.

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Original post by omnipotxnce
S = {2x^2 : x ∈ R and x^3 > −8}I need help with determining the inf (s) and the sup (s) and whether theyre in the set or not. Thanks in advance.

It is a good start to determine what are the allowed x values.
Reply 2
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omnipotxnce
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Original post by RDKGames
It is a good start to determine what are the allowed x values.


in interval form: S = { (-infinity, +infinity) union [-2,2)} or S = { (-infinity, +infinity) union [-2, +infinity)}
Original post by omnipotxnce
in interval form: S = { (-infinity, +infinity) union [-2,2)} or S = { (-infinity, +infinity) union [-2, +infinity)}

So which one is it?

It's just the solution to x^3 > -8

Also, its INTERSECTION and not UNION.
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Original post by RDKGames
So which one is it?

It's just the solution to x^3 > -8

Also, its INTERSECTION and not UNION.


okay so S = { (-infinity,+infinity) intersection [-2,2)} which gives inf (s) = -2, sup (s) doesnt exist and inf (s) is also in the set?

I struggle so much with real analysis and proof especially so apologies for asking continuous questions.
Original post by omnipotxnce
okay so S = { (-infinity,+infinity) intersection [-2,2)} which gives inf (s) = -2, sup (s) doesnt exist and inf (s) is also in the set?

I struggle so much with real analysis and proof especially so apologies for asking continuous questions.

Don't try to work out what S is yet, just work out the set of values s.t. x^3 > -8.
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omnipotxnce
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Original post by DFranklin
Don't try to work out what S is yet, just work out the set of values s.t. x^3 > -8.

Im not trying to work out S im trying to work out inf (S) and sup (S) but i dont know how
Original post by omnipotxnce
Im not trying to work out S im trying to work out inf (S) and sup (S) but i dont know how

The first thing is to work out the set of allowable values for x.

You will then need to work out what's actually in S so you can find the sup and inf.

The expressions you've been posting for S are wrong, but because you're trying to do it all at once and not posting working, it's impossible to tell what you're actually doing wrong.
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Original post by DFranklin
The first thing is to work out the set of allowable values for x.

You will then need to work out what's actually in S so you can find the sup and inf.

The expressions you've been posting for S are wrong, but because you're trying to do it all at once and not posting working, it's impossible to tell what you're actually doing wrong.

I literally dont know how to work out what the allowed x values are or how to work out whats actually in S or just how to do the question. My expressions were just in interval form to help me better read them but as youve said im wrong so i have nothing to go off. are u able to show me the steps needs to get the answers i need ?
Original post by omnipotxnce
I literally dont know how to work out what the allowed x values are or how to work out whats actually in S or just how to do the question. My expressions were just in interval form to help me better read them but as youve said im wrong so i have nothing to go off. are u able to show me the steps needs to get the answers i need ?

So, just to be clear: you're telling me you don't know how to find the set of values for which x^3 > -8?
Original post by DFranklin
So, just to be clear: you're telling me you don't know how to find the set of values for which x^3 > -8?

{(-2, + infinity)} ?

if this isnt right then yes i dont know how to do it
Original post by omnipotxnce
{(-2, + infinity)} ?

if this isnt right then yes i dont know how to do it

This is correct! So your set S is the equivalent to the values that 2x^2 gives when x is in the set (-2,+inf).

You can sketch the set of values of S to help you see the inf(S) and to talk about sup(S)
Original post by omnipotxnce
{(-2, + infinity)} ?

if this isnt right then yes i dont know how to do it

As RDK says, that's correct. Follow his advice in post#12
Original post by DFranklin
As RDK says, that's correct. Follow his advice in post#12


Original post by RDKGames
This is correct! So your set S is the equivalent to the values that 2x^2 gives when x is in the set (-2,+inf).

You can sketch the set of values of S to help you see the inf(S) and to talk about sup(S)

okay so would my S then be S = {(8, + infinity) intersection (-2, +infinity)}

with inf(s) = -2, and sup(s) = 8. inf(s) doesnt exist in the set whereas sup(s) does?

or am i skipping steps again ?
Original post by omnipotxnce
okay so would my S then be S = {(8, + infinity) intersection (-2, +infinity)}

with inf(s) = -2, and sup(s) = 8. inf(s) doesnt exist in the set whereas sup(s) does?

or am i skipping steps again ?

No that's not what S would be. Can you explain *why* you thought that's what S would be, because I've absolutely no idea what you've done here.
Original post by omnipotxnce
okay so would my S then be S = {(8, + infinity) intersection (-2, +infinity)}

with inf(s) = -2, and sup(s) = 8. inf(s) doesnt exist in the set whereas sup(s) does?

or am i skipping steps again ?

I think you are having a hard time working with the notation here.

The set S={2x2:x>−2}S = \{ 2x^2 : x>-2 \} contains the values of 2x22x^2 for x>−2x>-2.

Why don't you sketch y=2x2y=2x^2 for x>−2x>-2 ? This is a simple quadratic, The y-values it takes correspond to the set S.

The infimum would be the GREATEST lower bound for the y-values of this quadratic.
Original post by RDKGames
I think you are having a hard time working with the notation here.

The set S={2x2:x>−2}S = \{ 2x^2 : x>-2 \} contains the values of 2x22x^2 for x>−2x>-2.

Why don't you sketch y=2x2y=2x^2 for x>−2x>-2 ? This is a simple quadratic, The y-values it takes correspond to the set S.

The infimum would be the GREATEST lower bound for the y-values of this quadratic.

so from my sketch i got inf(s) to be 1 and sup (s) is unbounded ?
Original post by omnipotxnce
so from my sketch i got inf(s) to be 1 and sup (s) is unbounded ?

There is no upper bound, so sup(S) does not exist.

Check your infimum though. 1 cannot be a lower bound since if x=0.1 then 2x^2 = 0.02 which is less than 1.
Original post by RDKGames
There is no upper bound, so sup(S) does not exist.

Check your infimum though. 1 cannot be a lower bound since if x=0.1 then 2x^2 = 0.02 which is less than 1.

so infimum must be 0? as square function corresponds to the postives only so anything below zero becomes positive again
Original post by omnipotxnce
so infimum must be 0? as square function corresponds to the postives only so anything below zero becomes positive again

Yes the infimum is zero. Is it in the set S?

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