T/F continuous functions

[dapper spear]

given: f(x) > x for every x in [0,1],
T/F : if f is continuous in (0,1) => inf(f((0,1)) > 0

please help

[raw sedge]

!status

[dapper spear]

still trying to find a contradicting example, but cant seem to find one still.

[raw sedge]

might not be a contradiction then

[dapper spear]

how may I approach such proof? I thought about assuming towards a contradiction inf(f((0,1)) <= 0 but can't think of a way this will work

[raw sedge]

that's a good assumption to make for contradiction proof

First, in your words, what does $\inf(f(0, 1))$ represent?

[steep furnace]

SWR

[dapper spear]

The highest lower bound (the max number which bounds f from below)
meaning it has to be <= to every number f((0,1)) gives.

Is it correct?: if we assume inf(f((0,1))) = 0, f(x) can be undefined in x = 0. Therefore, the given information (f(x) > x for every x in [0,1]) still holds?

so this will make a contradiction to the whole sentence.
for example $\f(x) = x + 2^(-1/x)$

[steep furnace]

Yar
Compile Error! Click the reaction for more information.
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[dapper spear]

I tried

[raw sedge]

correct

[raw sedge]

First, $f((0, 1))$ is a set, you I think you mean $f((0, 1))={0}$. It's good practice to be consistent with notation here

[steep furnace]

SWR

[raw sedge]

I do not understand what you are trying to get at with $f(x)$ being defined or undefined at $x=0$. I don't see how whether it being defined or not makes any contribution to the proof

[steep furnace]

SWR

[raw sedge]

Do you remember what f((0, 1)) represents?

[dapper spear]

oh sorry, I ment inf(f((0,1))) = 0

[raw sedge]

ah

[raw sedge]

so many parentheses

brb 5 min

[dapper spear]

In other words:
When f(x) -> 0 (as x is approaching 0+) and f(x) = 0 is undefined, the inf(f((0,1))) = 0 and at the same time, f(x) > x for every x in [0,1]?

(and of-course f(x) has to be non-negative)

[raw sedge]

brb longer gotta take care of some stuff

[dapper spear]

No problem

[raw sedge]

@dapper spear I am back

"f(x) = 0 is undefined" it's better to say "$0\notin f([0, 1])$" or "$f(x)\ne 0$ for all $x$"

[steep furnace]

SWR

[dapper spear]

Thanks, so I got the right idea in your opinion?

[raw sedge]

No. I don't think so. Just trying to help with some of your wording

Okay

Let's try a proof by contradiction

Suppose $\inf f((0, 1))\le 0$. What can you infer from this?

[steep furnace]

SWR

[dapper spear]

that f's highest bound from below is smaller or equal then 0 in (0,1).
that every number that f(0,1) gives is bigger (or equal if min(f(0,1)) is there) number then 0.
Not entirely sure on the min part.

[raw sedge]

yeah you can't really use the min part

but you are close to a good idea

This is where you may need some $\varepsilon$-$\delta$ usage

[steep furnace]

SWR

[raw sedge]

Like, say $\inf f((0, 1))=0$, then, for any $\varepsilon>0$, there is some $x\in(0, 1)$ such that $f(x)=\varepsilon$

[steep furnace]

SWR

[raw sedge]

But, $f(x)>x$ for all $x\in (0, 1)$. So, if $f(x)=\varepsilon$, then it must be that $x<\varepsilon$

[steep furnace]

SWR

[dapper spear]

Right, but that isn't a contradiction yet as we only know that $\varepsilon>0$.
This can mean that $lim(f(x)) = 0$ as x is approaching 1- ?

[steep furnace]

Yar

[raw sedge]

hmmm

trying to think of what to do next

I know it needs continuity, but the exact play here is tricky

ultimately, you want to prove that $\inf f((0, 1))\in f((0, 1))$

[steep furnace]

SWR

[dapper spear]

but if $lim(f(x)) = 0$ as x is approaching 1- , is i even possible? as we only assume $\inf f((0, 1))=0$

[steep furnace]

Yar

[raw sedge]

why is $\lim_{x\to1^-} f(x)=0$?

[steep furnace]

SWR

[raw sedge]

How are you figuring that?

[dapper spear]

Because we inferred that $x<\varepsilon$

[steep furnace]

Yar

[raw sedge]

I think you misunderstood what I was saying here

oh

bruh

[dapper spear]

I think i understood what you did, but didn't get how can I use what we inferred to continue with the proof

[raw sedge]

I just thought of a counterexample

[dapper spear]

🛌

[raw sedge]

Try to think of a continuous function where $\lim_{x\to0^+}f(x)=0$ and $f(x)>x$ for all $x\in(0, 1)$

[steep furnace]

SWR

[raw sedge]

any old function will do.

[raw sedge]

@inner sierra fyi, there is a contradiction

[dapper spear]

$\f(x)=\root(x)$ for example

[steep furnace]

Yar
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dapper spear]

I was mistaken.
Maybe f(x) = root(x) + 1

no its 1
give me a second

[raw sedge]

it can be simpler than this

[dapper spear]

maybe f(x) = x/root(x)

Can't think of something simpler

[dapper spear]

maybe for this context just f(x) = 2x ?

[dapper spear]

Does it has anything to do with the function increasing and $x<\varepsilon$?

[steep furnace]

Yar

[raw sedge]

[raw sedge]

Can you think of ways to slightly alter this so that all of the requirements are met, but inf ends up being 0?

[dapper spear]

f(x) = |2x| ?

[raw sedge]

@dapper spear, sorry, I was eating

absolute value doesn't really change anything

f is defined in [0, 1], so |2x| is just 2x in that domain interval

[dapper spear]

Can't think on something simpler then:

Sorry, had to go for a bit

[raw sedge]

really, just $f(x)=2x$ for $x\in(0, 1]$, and then $f(0)=1$

[steep furnace]

SWR

[dapper spear]

but then in x=0 f(x)=x and not >

[raw sedge]

Notice how i defined f(0)=1

[dapper spear]

Oh, you are doing a function with cases
Do you think we can actually use something like that here?

[raw sedge]

yes

[dapper spear]

Thank you for your help

Very much appreciated