Proof check: continuous functions (General topology)

[mathmonkey]

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 $f$.

Homework Equations

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 $\overline{A}$.

 

[hedipaldi]

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[mathmonkey]

Hi hedipaldi,

Thanks for the response! Sorry, I forgot to explicitly mention this in the proof, but I believe that since the range $Y$ is a Hausdorff space by hypothesis, this implies that for two distinct points in $Y$, there exists disjoint open neighborhoods containing each point. This ensures that the pre-image of both of these open neighborhoods are also disjoint, except for the single point $x$.

 

[hedipaldi]

Yes,but there may be other accumulated points of A common to the two pre images.

 

[mathmonkey]

Hmm i see what you're saying. You mean their intersection on $A$ is empty, but they may still intersect at other points of $A'$ right?

How about this then? I can modify the argument to show that any subset of $A' - A$ cannot be open, where the contradiction will still hold since $g^{-1}(U) \cap h^{-1}(V) \subset A' - A$. And any nonempty subset of $A' - A$ cannot be open since its elements are all limit points of $A$ and any open set containing them must intersect $A$. Does this rectify the problem?

However, I'm confused on the part of your proof where you say "there exists $a_1,a_2 \in A, a_1 \in g^{-1}(U), a_2 \in h^{-1}(V)$, so $g(a_1) = g(a_2)$ is in $U \cap V$, contradiction to $U \cap V$." What do you mean by $g(a_1) = g(a_2)$? How did we know that $g(a_2) \in U$?

 

[hedipaldi]

since g(a1)=h(a2) and g(a1) is in U while h(a2) is in V,we have an element in the intersection of U and V.
I,mistakenly wrote g insted of h.

 

[hedipaldi]

your last argument for the proof seems to be right.