Given a function $f$ infinitely differentiable at a point $c$ does there exist a neighborhood of $c$ in which $f$ is infinitely differentiable?

Suppose $$f$$ is a real valued function defined on a subset of the reals and $$f$$ is infinitely differentiable at $$c$$.Then is it possible that there does not exist any neighborhood of $$c$$ in which $$f$$ is infinitely differentiable.Clearly $$c$$ cannot be an isolated point of any of the domains of $$f^{(k)}$$.But I cannot proceed further. Alternatively can we say that the set $$S(f) :=\{c : f$$ is infinitely differentiable at $$c\}$$ is an open set?

This is similar to the idea in Kavi Rama Murthy's answer. First, note that for $$k = 0, 1, 2, \dots$$, you can construct a function $$h_k : [-1, 1] \to \mathbb{R}$$ which is $$C^k$$ but is not $$(k+1)$$ times differentiable at any point: just take repeated antiderivatives (integrals) of a continuous nowhere differentiable function, e.g. the Weierstrass function. By finding a smooth function $$g_k$$ with $$g_k^{(i)}(-1) = h_k^{(i)}(-1)$$ and $$g_k^{(i)}(1) = h_k^{(i)}(1)$$ for $$i = 0, 1, \dots, k$$, you can construct $$f_k = h_k - g_k$$, which has $$f_k^{(i)}(-1) = f_k^{(i)}(1) = 0$$ for $$i = 0, 1, \dots, k$$ -- this is a $$C^k$$ function on $$\mathbb{R}$$ supported on $$[-1, 1]$$ which is not $$(k+1)$$ times differentiable anywhere on $$[-1, 1]$$.
Now, consider the function $$f : [-1, 1] \to \mathbb{R}$$ defined as follows. On $$(1/(n+1), 1/n)$$, we let $$f$$ be equal to a $$C^n$$ function supported in this interval which is nowhere $$(n+1)$$-differentiable on its support (similar to one constructed above), and we scale $$f$$ down sufficiently so that on this interval, $$|f^{(i)}| \leq e^{-(n+1)^2}$$ for $$i = 0, 1, \dots, n$$ (which is possible since these derivatives are all bounded). Finally, define $$f(0) = 0$$ and $$f(x) = f(-x)$$ for $$x < 0$$. It is clear that $$f$$ is $$C^n$$ on the intervals $$(-1/n, 0)$$ and $$(0, 1/n)$$, and thus it is $$C^n$$ on $$(-1/n, 1/n)$$ since $$|f^{0}(x)|, |f^{(1)}(x)|, \dots, |f^{(n)}(x)| \leq e^{-1/x^2}$$ near $$x = 0$$. Thus $$f$$ is infinitely differentiable at $$0$$. However, it is not infinitely differentiable on any neighborhood of $$0$$: within $$(-1/n, 1/n)$$ there are points where it is not $$(n+1)$$ times differentiable.