Magento CE 1.9.0.1

I'm interested in echoing out the product attribute set name on the catalog/category pages. This sounds like a strange thing to do but we use Magento's attribute set in a unique way for classifying products.

I was attempting to pull it from the Mage::getModel('catalog/category') feature in my page template but was unsuccessful.

I also tried to pull the name with the following code:

$attributeSetModel = Mage::getModel("eav/entity_attribute_set");
$attributeSetModel-&gt;load($product-&gt;getAttributeSetId());
$attributeSetName  = $attributeSetModel-&gt;getAttributeSetName();
echo "&lt;p&gt;" . $attributeSetName "&lt;/p&gt;";


Is there a way to get the name to display with their products on the frontend catalog pages? Thanks in advance.


Question: Prove that any real function defined on a real open interval has at most countably many simple discontinuities.
Answer: Let $E_n = \left\{ x \in \left( a , b \right) \mid \lvert \lim_{t \to x^+} f(t) - 
\lim_{t \to x^-} f(t) \rvert &gt; \frac{1}{n} \right\}$, the set of simple discontinuities whose jumps are greater than $\frac{1}{n}$. Then $E_1 \subset E_2 \subset\cdots$ and $\cup^\infty_{i=1} E_i$ is the set of all simple discontinuities. If all of the $E_n$ are countable, then $\cup^\infty_{i=1}E_i$ is countable.

Suppose $ x \in E_n $.

The left-hand limit exists, so

$$
\exists \delta^-_x &gt; 0 \colon t \in \left( x - \delta^-_x, x \right) \implies \lvert f(t) - \lim_{t \to x^-} f(t) \rvert &lt; \frac{1}{4n}
$$

Suppose $y \in \left( x - \delta^-_x, x \right)$ is a simple discontinuity. Then

$$
\exists \delta^-_y &gt; 0 \colon t \in ( y - \delta^-_y, y ) \implies \lvert f(t) - \lim_{t \to y^-} f(t) \rvert &lt; \frac{1}{4n} \\
\exists \delta^+_y &gt; 0 \colon t \in ( y, y + \delta^+_y ) \implies \lvert f(t) - \lim_{t \to y^+} f(t) \rvert &lt; \frac{1}{4n}
$$

Then $\exists z^- \in (y - \delta^-_y, y) \cap (x - \delta^-_x, x), z^+ \in (y, y + \delta^+_y) \cap (x - \delta^-_x, x) \colon$

$$
\lvert \lim_{t \to y^+} f(t) - \lim_{t \to y^-} f(t) \rvert \leq \\
\lvert \lim_{t \to y^+} f(t) - \lim_{t \to x^-} f(t) \rvert + \lvert \lim_{t \to y^-} f(t) - \lim_{t \to x^-} f(t) \rvert \leq \\
\lvert \lim_{t \to y^+} f(t) - f(z^+) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^+) \rvert + \lvert \lim_{t \to y^-} f(t) - f(z^-) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^-) \rvert \leq \\
\frac{1}{n}
$$

So that 

$$
\forall x \in E_n \exists \delta &gt; 0 \colon y \in (x - \delta, x )  \implies y \not \in E_n
$$

This says that for a jump at $x$, there exists a left-hand neighborhood around x such that the jumps in the neighborhood are smaller than the jump at $x$ (we could prove the same thing for a right-hand neighborhood, but that turns out to be unnecessary). For each $x$, choose a rational number $q \in \mathbb{Q}$ from $(x - \delta, x )$, e.g., 

$$
\frac{ \lceil (x - \delta) ( \lceil \frac{1}{\delta} \rceil + 1 ) \rceil + 1}{\lceil \frac{1}{\delta} \rceil + 1}
$$

Since the $(x - \delta, x)$ are disjoint, $g \colon x \mapsto q$ is an injection $g \colon E_n \rightarrowtail \mathbb{Q}$ (each $x$ maps to a different $q$). Since $\mathbb{Q}$ is countable, so is $E_n$.