Discussion
An equivalence relation ~ defined on X = [0, 1] by \(s \sim t\) iff either \(s = t\) or \(s, t \in \{0, 1\}\). Then quotient space X/~ (if [0, 1] is given the metric topology induced by modulus) is homeomorphic to :
Answer: D
The quotient is homeomorphic to the circle \(s^{1} = \{(x, y) : x^{2} + y^{2} = 1\}\).
The relation glues the two endpoints 0 and 1 together and leaves every other point alone. The map \(g(t) = (\cos 2\pi t, \sin 2\pi t)\) from [0, 1] onto the circle is continuous and identifies exactly 0 and 1. It therefore induces a continuous bijection from \(X/\sim\) onto \(S^{1}\). The quotient of a compact space is compact and the circle is Hausdorff, so this continuous bijection is a homeomorphism.
It is not [0, 1], [0, 1) or (0, 1): removing any single point from the circle leaves it connected, while removing an interior point from any interval disconnects it. [0, 1) and (0, 1) are also not compact, while the quotient is.
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