# Do these functions exist? [duplicate]

I created this question, but, I do not know the answer:

Is there a function $$f: \mathbb R \to \mathbb R$$ such that for every interval $$I \subseteq \mathbb R$$ we have $$f(I)=\mathbb R$$?

It seems to me that existence of such a function would violate almost everything that I know about analysis, but, on the other hand, it would be awesome if there is at least one such $$f$$, because it is known that there are many "pathological" examples in analysis. Also, many here are skilled in analysis more than I am, so I can expect an answer, curious about what it will turn to be.

## marked as duplicate by Kabo Murphy real-analysis 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(); } ); }); }); Jun 18 at 5:35

There's for instance Conway's base-$$13$$ function.