Is a Weierstrass curve a topological manifold? [duplicate]

Obviously a Weierstrass curve is not a smooth manifold, but it seems like a Weierstrass curve should be a topological manifold (which I now see is a suspicion supported by this post), since it is a continuous image of the real line.

That said, I am a particularly concerned about the locally Euclidean requirement. I know that a Weiestrass curve generally has a non-integer Hausdorff dimension (see here)--does that impact this requirement?

marked as duplicate by Moishe Kohan, José Carlos Santos general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 30 at 8:51

A Weierstraß curve $$W \subset \mathbb R ^2$$ is the graph $$G(w) = \{ (x,w(x) \mid \in J \}$$ of a Weierstraß function $$w : J \to \mathbb R$$ (which is continuous but nowhere differentiable). Here $$J$$ is an interval. The map $$i : J \to G(w), i(x) = (x,w(x))$$, is a homeomorphism. This is true for any graph of any continuous function $$f : X \to Y$$ between toplogical spaces.
Therefore $$G(w)$$ is a topological manifold. It is also a topological submanifold of $$\mathbb R ^2$$. See my answer to The graph of $f:\mathbb R^2\to \mathbb R$ can be embbeded in $\mathbb R^2$ or only in $\mathbb R^3$?.
Moreover, it can be given the structure of a differentiable manifold: Simply take the single chart $$i^{-1} : G(w) \to J$$. However, it is not a differentiable submanifold of $$\mathbb R ^2$$.