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In the book Simmons, George F., Introduction to topology and modern analysis, page no- 98, question no- 2, the problem is : Let $X$ be a topological space and a $Y$ be metric space and $f:A\subset X\rightarrow Y$ be a continuous map. Then $f$ can be extended in at most one way to a continuous mapping of $\bar{A}$ into $Y$.

I am trying to prove this way. Let $x_0\in \bar{A}-A$ and suppose that there is two extension $f$ and $g$ such that $f(x)=g(x)$ for $x\in A$. Now $f(x_0)\in \overline{f(A)}$ and $g(x_0)\in\overline {g(A)}$. So there exists a sequence $\{f(x_n)\}$ and $\{g(y_n)\}$ that converge to $f(x_0)$ and $g(x_0)$ respectively, where $x_n$ and $y_n$ belong to $A$ for all $n$. Then I am stuck!! Please help to complete the proof.

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The following is a well-known result in point-set topology.

Proposition. Two continuous functions $f, g \colon X \to Y$ from a topological space $X$ to a Hausdorff space $Y$, that coincide over a dense subset $D \subseteq X$, necessarly coincide everywhere.

Proof. Consider the set $$Z :=\{x \in X \, | \, f(x)=g(x)\} \subseteq X.$$ Then $Z$ is closed in $X$, since it is the preimage of the diagonal of $Y \times Y$ (that is closed because $Y$ is Hausdorff) via the continuous map $$h \colon X \to Y \times Y, \quad x \mapsto (f(x), \, g(x)).$$ On the other hand, by assumption $D \subseteq Z$ and so, since $D$ is dense in $X$, we obtain $$X = \bar{D} \subseteq \bar{Z} = Z,$$ that is $X = Z$ and the proof is complete.

Now we can get what you want from the Proposition above, because $A$ is dense in $\bar{A}$ and $Y$ is metric, hence Hausdorff.

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To follow your idea: Suppose we have $f_1$ and $f_2$ that are both continuous extensions of $f: A \to Y$ to $\overline{A}$. Let $p \in \overline{A}$ and so we have a net $a_i, i \in (I,\le)$ from $A$ such that $a_i \to p$.

The continuity of $f_1$ implies that $f_1(a_i) \to f_1(p)$.

The continuity of $f_2$ implies that $f_2(a_i) \to f_2(p)$ .

But the nets (as $f_1(a_i) = f_2(a_i) = f(a_i)$) are identical so must converge to the same point in the metric space $Y$ (limits of sequences are unique).

We conclude that $f_1(p) = f_2(p)$ As this holds for all $p \in \overline{A}$, $f_1 = f_2$ on $\overline{A}$.

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  • $\begingroup$ In general topological space does it always true that if a sequence $\{x_n\}$ converges to x then $f(x_n)$ also converges to $f(x)$, where $f$ is continuous function ? $\endgroup$ – chandan mondal Sep 12 '17 at 17:04
  • $\begingroup$ Yes@chandanmondal , only the reverse need not hold: preserving sequential convergence need not imply continuity. But continuous always implies sequential continuity. $\endgroup$ – Henno Brandsma Sep 12 '17 at 17:55
  • $\begingroup$ Now I understand. Thank you very much $\endgroup$ – chandan mondal Sep 12 '17 at 19:06

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