# For each continuous $g:X\to [0,1]$, $g(a_n)\to g(a)$, can we deduce $a_n\to a$?

Assume

• $(X,\mathcal T)$ is a Tychonoff space.
• $(a_n)$ is a sequence in $X$.
• $a\in X$.
• for each continuous function $g:X\to [0,1]$, $$g(a_n)\to g(a)$$

Is there an elementary proof for $$a_n\to a$$ ?

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If $a_n$ doesn't converge to $a$ then there exists an open neighbourhood $U$ of $a$ and a subsequence $b_n$ of $a_n$ such that $b_n$ lie outside $U$. Let $g:X\rightarrow \mathbb{R}$ be continuous such that $g(a)=0$ and $g(x)=1$ for all $x\notin U$ (here I am using the definition of Tychonoff space). Then $g(b_n)=1$ eventually and $g(a)=0$.