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First of all, i'm sorry that i don't know what the title should be for this question. Please edit the title if there is a better way to describe this question.

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Here's a definition from wikipedia;

Let $X,Y$ be topological spaces and $Y$ be an ordered set and $E\subset X$ and $f:E\rightarrow X$ be a function.

Then $\limsup_{x\to a} \triangleq \inf \{\sup\{f(x)\in Y|x\in E\cap V\setminus \{x\}\}\in Y|V \text{is open}, E\cap V\setminus \{a\} ≠ \emptyset, a\in V\}$.

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Now, let $X$ be a metric space and $Y$ be an ordered topological space and $f$ is a map from $E(\subset X)$ to $Y$ and $a$ be a limit point of $E$.

I have shown that "If there exist $\limsup_{x\to a} f(x)$, then $\inf_{\epsilon >0} (\sup \{f(x)\in Y|x\in E\cap B(a,\epsilon)\setminus \{a\}\})$ exists and it is equal to $\limsup_{x\to a} f(x)$."

However, i am not sure if the existence of converse is true.

Thank you in advance

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