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Let $u,v\in f(I)$ with $u<y<v$. Choose preimages $x_u,x_v\in I$, and let $a=\min (x_u,x_v)$, $b=\max (x_u,x_v)$. The defining property of a real interval ensures $[a,b]\subseteq I$. The restriction of $f$ to this closed interval is continuous, and its endpoint values are $u,v$ in one order or the other. The intermediate value theorem supplies $c\in [a,b]\subseteq I$ with $f(c)=y$. Thus every value between two image values lies in the image, provingThe continuous image of a real interval result does not require $I$ itself to be closed or bounded; only the segment between the chosen preimages is used. Empty and single-point images also satisfy the interval property.

A suitable continuous function isFor $t_k=\pi /2+2\pi k$ and $u_k=3\pi /2+2\pi k$,All these arguments lie in $(0,1]$. Its image is a real interval by the continuous image of a real interval result, and it is unbounded both above and below. Given any real $y$, choose two of the displayed points with function values on opposite sides of $y$, then apply the intermediate value theorem on the segment joining them. Hence the image is all of $\mathbb{R}$. A monotone example would not suffice here; the oscillation gives both directions of unboundedness near the same missing endpoint.