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**1. Homework Statement**

Let ##A \subset X##; let ##f:A \mapsto Y## be continuous; let ##Y## be Hausdorff. Show that if ##f## may be extended to a continuous function ##g: \overline{A} \mapsto Y##, then ##g## is uniquely determined by ##f##.

**2. Homework Equations**

**3. The Attempt at a Solution**

Just looking for some verification/feedback/comments on my proof. Not sure if the way I handled it was correct, or the best way to approach the problem. Any help would be greatly appreciated!

Proof:

Let ##g## and ##h## both be extensions of the continuous function ##f##. We wish to show that they are equal everywhere. We need not check for points in ##A## since both agree with ##f## everywhere on ##A##, so it is sufficient to consider only limit points of ##A##. First we show that if ##x \in A'##, but ##x \not \in A##, then the set ##\{x\}## is not open in ##\overline{A}##. For since ##x## is a limit point of ##A##, every open set containing ##x## intersects ##A##, so that this open set must contain more than one element.

Now, suppose ##g(x) \not = h(x)##. By hypothesis, there exists open sets ##U## containing ##g(x)## and ##V## containing ##h(x)## such that the two sets are disjoint. Since ##g## and ##h## are both continuous functions, ##g^{-1}(U)## and ##h^{-1}(V)## must be open in ##\overline{A}##. However, note that ##g^{-1}(U) \cap h^{-1}(V) = \{x\}##, which is a contradiction, since finite intersections of open sets must be open, and we have already shown that ##\{x\}## is not open. Hence, ##g(x) = h(x)## for all ##x \in A'##, so that ##g = h## on ##\overline{A}##.