Question: I came upon the following question (probably trivial...): 

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> Let $X$ be a topological space Suppose $N \subseteq M \subseteq X$ and $M$ is closed and discrete in $X$. Then is it true that $N$ is closed and discrete in $X$?
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My thoughts: I believe the statement is *true*. 

Let $a \in N \subseteq M$, Then exists nhood $U\_a \cap N = U\_a \cap M = \{ a \}$. Thus $N$ is discrete. Suppose $N$ has an accumulation point; exists $x \in X$ such that $\{ x\_n \} \subseteq N \subseteq M$ with $x\_n \rightarrow x$. Then $x$ is also an accumulation point of $M$. As $M$ is closed, $x \in M$, contradicting discreteness. Hence, $N$ has no accumulation point, and is therefore closed.

Answer: 
Here is another ways of visualizing the problem. First, we arrange $16$ blue balls and $10$ green balls so that no two of the green balls are consecutive. Then we label the balls. 

Line up $16$ blue balls, leaving spaces between successive balls and at the ends of the row. There are $17$ such spaces, $15$ between successive blue balls and two at the ends of the row.
$$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$
To ensure that no two of the ten green balls are adjacent, we choose $10$ of these $17$ spaces in which to insert a green ball, which can be done in 
$$\binom{17}{10}$$ 
ways. Next, we label the balls from left to right. The letters on the green balls are the desired subset of the alphabet in which no two of the selected letters are consecutive. For instance, 
$$\color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$
corresponds to the selection $\{A, C, G, J, L, N, R, T, V, X\}$.
A [similar question](https://math.stackexchange.com/questions/490405/choosing-numbers-without-consecutive-numbers) was asked a while ago, where $r$ non-consecutive numbers had to be selected from $n$ numbers. Basically, the problem at hand can be converted to a [stars-and-bars problem](https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)) with $11$ slots and $10$ bars, where the bars represent the indices of the selected letters. The remaining 16 letters must be divided among the $11$ slots, with at least one letter in each of the $9$ slots in the middle. Once one letter is placed in each of these slots, there are $26 - 10 - 9 = 7$ letters left. As such, the number of ways to select $10$ non-consecutive letters from the alphabet equals:

$${7 + 10 \choose 10} = {17 \choose 10} = 19448$$