How are the properties satisfied in this induction proof? I have some notes on the topic of the Principle of Induction (POI) from the perspective of the Well-Ordering Principle (WOP). The following claim has just been proved:
Claim: (Principle of Induction)
Suppose that $T ⊆ \mathbb N$, and that 
Property 1 $0 ∈ T$, and
Property 2 for every natural number $n$, if $n−1∈T$  then $n∈T$. 
Then $T = \mathbb N$.
The notes then go on to discuss how to use the result proven above, i.e the POI, in proofs. The example given is the following: 
Claim: 
$$ 1+2+...+n = \frac{1}{2}n(n+1) \tag{1} $$ 
Proof: 
For each integer $n$, either the equality $(1)$ holds, or it does not. Let $T$ be the set of those $n$ for which it is true:
i.e
$T = \{n ∈ \mathbb N: $ equality $(1)$ holds$\} $.
We check easily that $(1)$ holds for $n=1$, so $T$ has Property 1. 

This point of the proof is where one of my difficulties in understanding lies. Above, 'Property 1' is defined as $0∈T$ but here it says that $1 ∈ T$ is satisfying this property.

Now we check that it has Property 2. If $(1)$ is true for some integer $n$ then using the truth of $(1)$ for $n$, we get 
$$1+...+n+(n+1) = \frac{1}{2}n(n+1) + (n+1)$$

The proof goes on to show that the following is true, and I have skipped these steps.

$$1+...+n+(n+1) = \frac{1}{2}(n+1)(n+2)$$
So if $(1)$ holds for $n$ then it holds for $n+1$. Thus, $T$ has Property 2 as well, and therefore by the POI, $T$ is all of $\mathbb N$. In other words, $(1$) holds for all $n∈\mathbb N$.

I also have difficulties with understanding how this second property is satisfied. I interpreted Property 2 as being equivalent to: $$ n-1 ∈ T \implies n ∈ T .$$ Is it just that this is equivalent to saying that $$ n ∈ T \implies n+1 ∈ T ,$$ but just for some different starting point of $n$?

 A: Yes on both accounts.  
Specifically, there is always the issue as to whether the natural start with $0$ or $1$ ... but in practice, it is actually more important to know for which set of numbers you want to prove the claim. That is, sometimes we want to prove that some property is held by all whole numbers 0 and up. Other times we want to prove it for the numbers 1 and up (because it doesn't hold for 0, or is ill-defined for 0).  But other times, we may want to prove it only for the numbers, say, 3 and up.  So, in practice we just have our base case start with whatever the first number is of the set of numbers for which we want to prove the property.
And second, yes, instead of going from n to n+1, we can go from n-1 to n ... as long as we're careful! For example, if you want to prove that something is true for all numbers 1 and up (as in your example), then we can use as a base case n=1, but if we then want to use the method of going from n-1 to n as our step, we have to assume that the n that we pick in the step is at least 2, rather than 1.  And having to assume that could make all the difference. That is, if you're not careful, and assume that n could be 1, mistakes could take place.  Likewise, if you use the method og going from n to n+1, then after the base case of n=1, you need to assume that $n \ge 1$, rather than, say, $n \ge 2$
But yes, in general, there are many, many variants of induction. But as long as as you're careful and know the basic idea behind it, you can adopt your specific formalization to whatever problem you are dealing with.
P.s .. for for your example, you could include $n=0$, as the sum of all numbers of $1$ through $0$ can reasonably be defined as $0$, and as $\frac{0\cdot (0+1)}{2}$ is also $0$
A: I am assuming that $\mathbb{N}=\{0,1,2,\ldots\}$.
The statement of property $2$ doesn't make sense, because if $n=0$, then $n-1\notin\mathbb N$ and therfore $n-1$ cannot possibly be an element of $T$. My guess is that property $2$ is: if $n\in\mathbb{N}\setminus\{0\}$ is such that $n-1\in T$, then $n\in T$.
If I am right, then property $2$ is indeed equivalent to$$(\forall n\in\mathbb{N}):n\in T\implies n+1\in T.$$
