what are limit points of $X \times \mathbb{N}$? Let $X=\{0,1,2 \}$ has order topolgy and $\mathbb{N}$ has natural order topolgy. Suppose $X \times \mathbb{N}$ has dictionary order topology then what are limit points of $X \times \mathbb{N}$ ?
Answer:
\begin{align*}X \times \mathbb{N} =& \{(0,1), (0,2), \cdots, (0,n), \cdots \\ & (1,1), (1,2), \cdots, (1,n), \cdots \\ & (2,1), (2,2), \cdots, (2,n), \cdots \}\end{align*}
Now let $<$ be the dictionary order, then $(x,y)<(a,b)$ if either $x<a $ or $x=a \ and \ y<b$. 
Any open set in $X \times \mathbb{N}$ has the form $U=\left((a,b), (a',b') \right)$, where $(a,b) < (a',b')$. 
If $(\alpha, \beta) \in U$, then $ (a,b)<(\alpha, \beta)<(a',b')$.
Then take the point $(0,3)$ and its nbd $U=((0,2),(0,4))$, clearly $(0,3) \in U$ but $U \cap ((X \times \mathbb{N}) \setminus \{(0,3)\})=\phi$. 
Hence $(0,3)$ is not a limit point.
Similarly we can show that no point is limit point.
So the set $X \times \mathbb{N}$ has no limit point.
Am I right?
 A: Your argument for $(0,3)$ works for $(m,n) \in X \times \mathbb N$ as long as $n \ge 2$, and a similar argument also works for $(0,1)$.
However, that argument does not work for $(1,1)$ or for $(2,1)$. 
To see why, consider a basis element $U$ of the topology that contains $(1,1)$, and so $U$ has the form $((a,b),(a',b'))$ where $(a,b) < (1,1) < (a',b')$. Since $(a,b) < (1,1)$ it follows that $a=0$. Thus the entire sequence of the points $(0,b+1), (0,b+2), (0,b+3),...$ is in the set $U$, and $(1,1)$ is a limit of that sequence. Since this is true for all choices of $U$ it follows that $(1,1)$ is a limit point of $X \times \mathbb N$. By a slightly more complicated argument you should also be able to prove that $(2,1)$ is a limit point.
By the way, just to correct a minor error, the elements $U = ((a,b),(a',b'))$ form a basis for the topology on $X \times \mathbb N$ as I said, however it is not true to say that any open set has that form. In general an open set can be expressed as a union of some collection of basis elements, but that collection might have more than just one basis element. 
