Here's some cute continuity problems:
- if two continuous functions $f,g$ agree on a dense set $D$,
  then they are equal.
  Pf:
   Most correctly, this works whenever the target is Hausdorff:
   ($f,g\from X \to Y$)
   then the set on which they agree is closed, as it's the
   pullback of the diagonal, which is closed by Hausdorff,
   and the closure of a dense set is the whole space.
   In a quip: ``equalizers (of maps to Hausdorff) are closed''.
   (indeed, are cofibrations, I think...)

  More concretely, given $x\in X$, take a sequence (or net)
  $x_n \to x$, where $x_n \in D$. Then $f(x_n)=g(x_n)$,
  and $f(x)=\lim f_n(x) = \lim g_n(x) = g(x)$, by continuity.

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- If $f,g\from \bR \to \bR$ are continuous functions that agree
  outside a set of measure zero, then they are equal.

  Pf 1: measure zero sets have empty interior (since open intervals
  have positive measure), so the compliment is dense:
  apply the previous problem.

  Pf 2: Apply FTC: $f(x)=F'(x)=G'(x)=g(x)$:
  by continuity (so $f=F'$) and $F=G$ b/c they agree
  off a set of zero measure, which doesn't affect integrals.

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- a locally constant function $f$ on an interval is constant

  Pf 1:
   Most correctly, a locally constant function
   from a connected space to a Hausdorff space
   is constant: use the open set defn of continuity.

  Pf 2:
   Try doing it using only IVT:
   Here's one proof:
   a locally constant function takes on only
   -countably- many values:
   (roughly, each value that the function takes,
    it takes on an open set, which contains a rational point,
    so the image of $f$ is the same as the image of
    $f$ restricted to $\bQ$, which is countable)
   Now if $f(x) \neq f(y)$, by IVT it must take on all of the
   uncountably many values in between, which it can't,
   so $f(x) = f(y)$ for all choices of $y$ and $f$ is constant.

  Yeah, this is a bit long -- is there a more straightforward
  way, just using IVT?

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