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I am working from Munkres' Analysis, and I've converted his definitions slightly to make them easier to compare. In the table below, you can fill in the blanks in the top row with words from the lower rows to form either definition:

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It is not clear to me what motivates some of the choices when it comes to 'filling in the blanks'. My biggest concern is the last two blanks. The 2nd blank is essentially discussed in my old question here:

Why not define 'limit' to include isolated points?

And while I roughly understand the response there (letting in isolated points means that functions can approach infinitely many limits at isolated points), when I consider changes to the last two columns, I find myself also considering changes to the 2nd.

My hope is that someone can construct simple examples for each column (in as few dimensions as possible!) which motivate the choice, while somehow dealing with the interconnection problem wherein choice in one column affects choice in another...

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The $x$ in the last column of the second row should be $a$? – joriki Aug 6 '11 at 11:42
whoops! Yes, it should. – Quine42 Aug 8 '11 at 4:09

First, note that by the definition of the subspace topology it wouldn't make a difference in the first row if you replaced $A$ by $X$ in the penultimate column and inserted $\cap A$ in the last column instead. So the difference between the rows in the last two columns is only that $a$ is being excluded. Excluding $a$ in the first row wouldn't make a difference since $f(a)$ is in $V$ anyway, so the only question remaining about these two columns is why $a$ is excluded in the second row. Not excluding it would complicate talking about the limiting behaviour of functions, e.g. you couldn't remove removable discontinuities by defining a new function using the limits of the function, but would have to explicitly exclude the point of discontinuity in each case. (I don't know if there are further reasons.)

We might still ask why we don't use $X$ in the penultimate column without $\cap A$ in the last column. This is because otherwise there won't be any suitable open sets if $x$ is on the boundary of $A$ but in the interior of $X$. For example, consider $X=Y=\mathbb R$ and $A=[0,1]$, and let $f(x)=0$. If we apply $f$ directly to $U$ without intersecting with $A$, it has to be a subset of $A$, but there are no subsets of $A$ containing $0$ that are open in $X$, so this function wouldn't be continuous if we required $U$ to be open in $X$.

Regarding the third column, I don't think there's much to explain; those are simply the points we're interested in, and they must be in $V$ because otherwise there's no reason to expect the rest of the definition to have anything to do with them.

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