Support of an exponential family in canonical form

Suppose $$Y$$ is a random variable in the exponential family, with pmf/pdf $$f(y) = \exp\left[\sum_{j=1}^{s}\theta_jT_j(y)-B(\theta)+c(y) \right]$$ for $$y \in \Omega$$ (the support of $$Y$$), and where $$\theta = \begin{bmatrix} \theta_1 \\ \theta_2 \\ \vdots \\ \theta_s\end{bmatrix}$$ is the canonical parameter.

I wish to show that

$$\Omega = \{y:\exp[c(y)] > 0\}\text{.}$$

where $$\Omega$$, the support of $$Y$$, is defined by $$\{y: f(y) > 0\}$$.

The Google searching I've found either just mentions this as a fact, or states that it's obvious. And I don't see why it is.

If we set $$f(y) > 0$$, we obtain $$\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]\exp[-B(\theta)]\exp[c(y)] > 0$$ or $$\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]\exp[c(y)] > 0\text{.}$$ The problem is, though, how does this imply that $$\exp[c(y)] > 0$$? Why is it that we ignore $$\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]$$?

• Can you find a number $x$ such that $\exp\{x\} \leq 0$? – jbowman Oct 3 '18 at 1:24
• @jbowman No, as long as $x \in \mathbb{R}$ (not allowing for $-\infty$). – Clarinetist Oct 3 '18 at 1:30
• So $\exp[c(y)] > 0$ because it can't be otherwise. – jbowman Oct 3 '18 at 1:34
• @jbowman I was doubting myself when I thought of something similar, because it seemed too simple... thanks! – Clarinetist Oct 3 '18 at 1:54
• Been there, done that! – jbowman Oct 3 '18 at 1:59