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This question has been asked before but my proof is different from what I have seen and so I would to ensure that it is correct.

Problem

Let $A \subset X$ and let $f:A \to Y$ be continuous, where $Y$ is Hausdorff. Assume that $f$ is extended to a continuous function $g: \overline{A} \to Y$. Show that $g$ is uniquely determined by $f$.

Proof:

Let $g_1$ and $g_2$ be such extensions and let $x \in \overline{A}$. Assume $g_1(x) \neq g_2(x)$ and let $V_1$ and $V_2$ be disjoint nbhds of $g_1(x)$ resp. $g_2(x)$. This is possible since $Y$ is Hausdorff.

Since $g_1$ and $g_2$ are continuous, there exists nbhds $U_1$ and $U_2$ of $x$ such that $g_1(U_1) \subset V_1$ and $g_2(U_2) \subset V_2$. Since $x \in \overline{A}$, it follows that $S = U_1 \cap U_2 \cap A \neq \emptyset$. Since $g_1 = g_2$ on $A$, we have that $g_1(S) = g_2(S)$.

Since $g_1(S) \subset g_1(U_1) \subset V_1$ and $g_2(S) \subset g_2(U_2) \subset V_2$, it follows that $g_1(S) = g_2(S) \subset V_1 \cap V_2 = \emptyset$, which is a contradiction. Hence $g_1 = g_2$ on $\overline{A}.$

JustANoob
  • 1,659

1 Answers1

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Alternative approach: Suppose $g_1, g_2$ be two continuous extension of $f$ on $\overline A$.

Then ${g_1|}_{A}=f=g_2|_{A}$

$S=\{a\in \overline A:g_1(a) =g_2(a) \}$ , then $S$ is closed set.

$A\subset S\subset \overline A$

$\overline A \subset S\subset \overline A$

Hence $S=\overline A$ and $g_1=g_2$ on $\overline A$.

Sourav Ghosh
  • 12,997