Linear system stability criteria Suppose we have a closed-loop system $H(s)=\frac{A(s)}{1+A(s)f(s)}=\frac{A(s)}{1+T(s)}$. I've seen the stability of the system stated a couple ways:

If $H(s)$ has any poles in the RHP, then it is unstable since the system response contains a term of the form $e^{s_{RHP}t}$, where $S_{RHP}$ has a positive real part and so will grow without bound over time.
If $1+T(s)=0$ for any $s$ (or equivalently $T(s)=-1$ for some $s$) then the system is unstable  since there exists some $s$ such that the transfer function blows up.

What confuses me is how these definitions are supposed to be equivalent. In particular, the second definition seems to say that if $H(s)$ has any poles, then the system is unstable. Why is this the case? Take for example a simple low-pass filter, $H(s)=\frac{1}{1+RCs}$, which has a pole at $s=-1/RC$. This pole is in the LHP and so by (1) is stable, however (2) seems to consider it unstable.
What is the relationship between (1) and (2)? In what situations, if any, are they equivalent/does one imply the other?
Only the first criterion is correct when referring to input-output stability. The second criterion is just a way to compute the poles, as mentioned in a comment. All realizable continuous-time systems using lumped elements have poles, which doesn't contradict the stability requirement.
Poles in the right half-plane (RHP) make a causal system unstable. Note that if we let go of the causality requirement, RHP poles don't cause instability but they make the system non-causal.
Note that poles on the imaginary axis cause transients that don't die out, which is undesirable in almost all practical applications. So in practice you want all poles to lie on the open left half-plane, i.e., $\textrm{Re}\{s_{\infty,k}\}&lt;0$, where $s_{\infty,k}$ is a pole of the system.