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Let $f,g:X\rightarrow Y$ be continuous where $Y$ is Hausdorff. Prove that if $f(x)=g(x)$ for each $x$ in a subset $A$ of $X$, then $f(x)=g(x)$ for each $x$ in $\overline{A}$.

Can anyone help me, please?

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Hint: Let $y\in \bar{A}$ be so that $f(y)\ne g(y)$. The space $Y$ being Hausdorff, implies there exist open sets $U_1$ and $U_2$ such that $$f(y)\in U_1,\;g(y)\in U_2\text{ and }U_1\cap U_2=\varnothing.$$

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open sets in $X$ containing $y\in \bar{A}$, therefore $$f^{-1}(U_1)\cap A\ne \varnothing\text{ and }f^{-1}(U_2)\cap A\ne \varnothing$$ Now ...

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The core idea is that if $X$ is Hausdorff then the diagonal $\Delta \subset X \times X$ is closed (this is a basic fact you should be able to prove).

Observe you can define the function $(f,g)\colon X\rightarrow X \times X$ by $(f,g)(x)=(f(x),g(x))$ which is clearly continuous, thus $\{x \in X : f(x)=g(x) \}=(f,g)^{-1}(\Delta)$ is closed and this gives you the proof of the statement.

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  • $\begingroup$ But in my question, $X$ is not necessarily Hausdorff. $\endgroup$
    – Kumar
    Feb 17, 2020 at 17:09
  • $\begingroup$ Sorry, I made some confusion. My argument still works perfectly if you replace the target of $f,g$ with $Y$: this space is Hausdorff so $\Delta \subset Y \times Y$ is still closed, hence its preimage via $(f,g)$ is closed. $\endgroup$
    – N.B.
    Feb 17, 2020 at 17:17

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