Does first countable imply equivalence of sequential and limit point compactness? Steen and Seebach say that: If a topological space $X$ is first countable, sequentially compactness is equivalent to limit point compactness in $X$. 
Take $X=\mathbb{N}\times\{0,1\}$, where $\mathbb{N}$ has the discrete topology and $\{0,1\}$ has the indiscrete topology. This space is first countable since the topologies on $\mathbb{N}$ and $\{0,1\}$ are first countable. Thus it should satisfy the statement in the previous paragraph. 
However, if one takes a sequence like $\{(n,0)\}$ in $X$, there are no convergent subsequences for limit points like $(1,1)$. In other words, one can't find a subsequence of $\{(n,0)\}$ that converges to $(1,1)$, even though $(1,1)$ is a limit point of the sequence. This seems to violate the statement in the first paragraph.   
I don't see where I'm making the error with this example. Any help is appreciated.
 A: You seem to have some confusion about the definitions.  Steen and Seebach say that for first-countable spaces, sequential compactness is equivalent to countable compactness, not limit point compactness.  You claim that their definition of countable compactness is what others call limit point compactness but this is false.  In particular, one of the equivalent conditions for countable compactness they give is:

Every sequence has an accumulation point.

This is not the same as limit point compactness, which means instead that

Every infinite set has a limit point.

Of course, limit point compactness can be checked just on countable sets since every infinite set has a countably infinite subset, but the key distinction is that an accumulation point of a sequence is not the same thing as a limit point of the set of terms of the sequence.  In particular, $x$ is an accumulation point of a sequence $(x_n)$ if for each neighborhood $U$ of $x$, there exist infinitely many $n$ such that $x_n\in U$.  In contrast, $x$ is a limit point of the set $\{x_n\}$ if for each neighborhood $U$ of $x$, there exists $n$ such that $x_n\neq x$ and $x_n\in U$.  In your example, $(1,1)$ is a limit point of the set $\{(n,0):n\in\mathbb{N}\}$ but is not an accumulation point of the sequence $(x_n)$ with $x_n=(n,0)$.
So, your space is indeed first countable, limit point compact, and not sequentially compact.  This does not contradict Steen and Seebach's statement though, since your space is not countably compact.
