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Can we classify all uniformly continuous functions from $X=S^1$-$\{(0,1), (0,-1)\}$ (circle -two points) to $1$-dim Euclidean space $\mathbb R$.

I feel that functions which take both the arcs of the circle to a fixed interval (same interval ) of $\mathbb R$ are uniformly continuous and these are the only uniformly continuous functions.

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It sounds like you are saying that a function $f: X \rightarrow \mathbb{R}$ is uniformly continuous if and only if it is bounded. Half of this is correct:

If $X$ is a metric space, then a uniformly continuous function $f: X \rightarrow \mathbb{R}$ is Cauchy continuous -- i.e., sends Cauchy sequences to Cauchy sequences, and thus extends to a uniformly continuous function on the metric completion $\overline{X}$. In this case $\overline{X} = S^1$, which is compact, so any uniformly continuous function on $X$ must be bounded.

But this is not sufficient, because not every bounded function on $X$ extends continuously to $S^1$. Indeed, topologically $X$ is just a disjoint union of two open intervals, so a function is continuous on $X$ iff its restiction to the left hand piece and the right hand piece are both continuous. So we could take for instance $f$ on the left hand piece to be the $y$-coordinate and $f$ on the right-hand piece to be identically zero, and this function will not extend continuously to $S^1$.

Note that extendability to $S^1$ is sufficient as well as necessary for uniform continuity on $X$, since any continuous function on a compact metric space is uniformly continuous and any function which is uniformly continuous on a space is uniformly continuous on each subspace of that space. So I think "$f$ extends to a continuous function on $S^1$" is the answer to your question; I would be (mildly) surprised if there is some other, better way to say it than this.

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