What is the flaw in the statement about T and how can it be corrected?

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Homework Statement



Let m be a natural number. Find the flaw in the statement below. Explain why the statement is not valid, and change one symbol to correct it.

"If T is a set of natural numbers such that 1) m \in T and 2) n \in T implies n+1 \in T, then T = {n \in N : n ≥ m}

Homework Equations



Dunno.

The Attempt at a Solution



Part 2) of the if statement tells us that T is an infinite set. I'm not sure exactly how 1) and 2) are connected. Hmmmm ...

Help me get started.
 
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To get started think about this. Is m-1 in T?
 
Dick said:
To get started think about this. Is m-1 in T?

Hmmm ...

T is going to look something like {k, k+1, k+2, ...}, where k≥1 is an integer. That's basically what the second condition tells me.

m is some element in T. That's all I know about m. Could m-1 be in T? As long as m>k.
 
So is their equation for T correct?
 
Jamin2112 said:
Hmmm ...

T is going to look something like {k, k+1, k+2, ...}, where k≥1 is an integer. That's basically what the second condition tells me.

m is some element in T. That's all I know about m. Could m-1 be in T? As long as m>k.

Ok, so you don't know if m-1 is in T. On the other hand, m-1 is definitely NOT in [m,infinity). That suggests that T and [m,infinity) are not necessarily the same thing.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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